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CBSE Class 11 Chemistry Notes For Chapter 8 Redox Reactions Introduction

Chemical reactions, according to their nature, are divided into different classes, such as combination reactions, decomposition reactions, elimination reactions, polymerisation reactions and many other types.

Oxidation-reduction or redox-reaction is a class of chemical reactions in which oxidation and reduction occur simultaneously. A large number of chemical and biological reactions belong to this class. Some processes that are associated with oxidation-reduction reactions are the rusting of iron, the production of caustic soda and chlorine by electrochemical method, the production of glucose by photosynthesis, the generation of electric power by battery, fuel cell, etc.

Redox Concept Reactions According To Electronic

Oxidation reaction:

A chemical reaction in which an atom, an ion or a molecule loses one or more electrons is called an oxidation reaction. An atom, ion or molecule is oxidised by loss of electron (s).

Examples: Oxidation reactions involving—

Loss of electron(s) by an atom:

Generally atoms of metallic elements such as Na, K, Ca, etc., undergo oxidation by losing electron(s), thereby producing positive ions.

⇒ \(\mathrm{Na} \longrightarrow \mathrm{Na}^{+}+e ; \mathrm{K} \longrightarrow \mathrm{K}^{+}+e ; \mathrm{Ca} \longrightarrow \mathrm{Ca}^{2+}+2 e\)

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Loss of electron(s) by a cation:

Some cations such as Fe²⁺, Sn²⁺, Cu⁺, etc., undergo oxidation by losing one or more electrons, thereby forming higher charges.

Fe²⁺→ Fe³⁺ + e; Sn²⁺ → Sn⁴⁺+ 2e

Cu²⁺ → Cu²⁺ + e

Loss of electron(s) by an anion:

Anions such as I^– and Br^– ions oxidise to neutral atoms or molecules by losing electron(s).

I₂^– →I₂ + 2e;2Br^– — Br₂ + 2e

Loss of electron(s) by a molecule:

Neutral molecules such as H₂, H₂O₂ and H₂O oxidise to cations by losing one or more electrons

H₂ → 2H+ + 2e;  H₂O₂→ O₂ + 2H⁺ + 2e

H₂O→ ½ O₂ + 2H⁺+ 2e

CBSE Class 11 Chemistry Notes Chapter 8 Redox Reactions

Reduction reaction:

A chemical reaction in which an atom, an ion or a molecule gains one or more electrons is called a reduction reaction. An atom, ion or molecule is reduced by the gain of electron(s).

Examples: Reduction reactions involving—

Gain of electron(s) by an atom:

Atoms of different elements in particular, such as, atoms of chlorine, bromine, oxygen and other non-metals are reduced to anions by gaining electron(s).

Cl + e → Cl^–

Br+ e→ Br^–

O+ 2e — O^2-

Gain of electron(s) by a cation:

Cations such as H⁺, Fe²⁺, Fe³⁺, Cu²⁺ etc., are reduced to neutral atoms or cations with lower charges by accepting electron(s)

H⁺+ e → H;  Fe²⁺+2e → Fe

Fe³⁺+ e → Fe²⁺ ; Fe³⁺+ 3e  → Fe

Cu²⁺+ 2e →Cu ; Cu²⁺+ e  → Cu⁺

Gain of electron(s) by a molecule:

Neutral molecules such as Cl₂, O₂, H₂O₂ etc., are reduced by gaining one or more electrons.

Cl₂ + 2e → 2CF ; O₂  + 4H⁺ + 4e→2H₂ O

H₂O₂ + 2H+ + 2e →  2H₂ O

Oxidant and reductant in light of electronic concept

Oxidant:

In a redox reaction, the species that itself gets reduced by accepting electron(s) but oxidises other substances is called an oxidant or oxidising agent. So, an oxidising agent is an electron acceptor. The greater the tendency of a substance to accept electrons, the greater the oxidising power it possesses.

Examples:

Oxygen (O₂), hydrogen peroxide (H₂O₂), halogens (F₂, Cl₂, Br₂, I₂), nitric acid (HNO₃), potassium permanganate (KMnO₄), potassium dichromate (K₂Cr₂O₇), sulphuric acid (H₂SO₄), etc.

Reductant:

In a redox reaction, the species that itself gets oxidised by losing electron(s) but reduces other substances is called a reductant or a reducing agent. So, a reducing agent is an electron donor.

The substance has a high tendency to lose electrons and acts as a strong reducing agent. Alkali metals (Na, K, Rb, Cs, etc.,) of group-IA of the periodic table show a strong tendency to lose electrons and hence behave as powerful reducing agents.

Examples: Hydrogen (H₂), hydrogen sulphide (H₂S), carbon (C), some metals (Na. K, Fe. Al, etc.), stannous chloride (SnCl₂), sulphur dioxide (SO₂), oxalic add (H₂C₂O₄), sodium thiosulphate (Na₂S₂O₂), etc.

According to an electronic concept;

• Oxidation involves the loss of one or more electrons. Reduction involves the gain of one or more electrons.
• Oxidants are electron acceptors. Reductants are electron donors.

Identification of oxidants and reductants with the help of electronic concept

Reaction: 2KI(aq) + Br₂(l)→2KBr(aq) + l₂(s)

The reaction can be represented in ionic form as

2K⁺ (aq) + 2I (aq) + Br₂(l)→ 2K^+- (aq)+ 2Br^–(aq) + I₂(s)……………………..(1)

This equation shows that in the reaction, the K⁺ ion does not undergo any change. The only function that it does In the reaction is to balance the charge. So, the K⁺ ion only acts as a spectator ion in the reaction.

Redox Reactions Class 11 Chemistry Notes

Hence, the net ionic equation of the reaction is-

2I(aq) + Br₂(l)→ I₂(s) + 2Br^–(aq)

Equation(2) shows that the I^– ion produces I₂ by losing electrons, while Br₂ forms Br^– ions by gaining electrons. 1 lenco, In this reaction, the conversion of I^– into

[2l^–(aq) → l₂(s) + 2e] is an oxidation reaction. On the other hand, the conversion of Br₂ into Br^– →2Br (l)] is a reduction reaction. Thus, in this reaction, ion, i.e., Kl is the reductant and Br₂ is the oxidant.

Oxidation-reduction occur simultaneously

Neither an oxidation reaction nor a reduction reaction can occur alone. Oxidation and reduction reactions are complementary to each other. In a reaction system, if a reactant loses an electron, then there must be another reactant In the system that will gain the lost electron. Thus, oxidation and reduction reactions must occur together.

In a redox reaction, the reducing agent gets oxidised by losing electron(s), while the oxidising agent gets reduced by accepting the lost electron(s). For example, Metallic zinc reacts with copper sulphate in a solution to form metallic copper and zinc sulphate.

⇒ \(\mathrm{Zn}(s)+\mathrm{CuSO}_4(a q) \rightarrow \mathrm{ZnSO}_4(a q)+\mathrm{Cu}(s) \downarrow\)

In an aqueous solution, CuSO₄ and ZnSO₄ exist almost completely In dissociated state. So, the reaction can be expressed in the form of an ionic equation as

⇒ \(\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)\)

In this reaction, the Zn atom loses two electrons, forming the Zn^2- ion (oxidation). On the other hand, the Cu²⁺ ion accepts these two lost electrons to produce Cu^– atom (reduction)

In this reaction, the Zn-atom can lose electrons only because the Cu²⁺ ion present in the reaction system accepts those lost electrons. Alternatively, the Cu²⁺ ion can get reduced by accepting electrons only because the Zn-atom present in the reaction system lose electrons.

So, in a reaction, when a substance undergoes oxidation, another substance present in the reaction system undergoes reduction. Tints it can be said that oxidation and reduction occur simultaneously.

Half-reaction

Any redox reaction consists of two half-reactions, one is for an oxidation reaction and the other is for a reduction reaction Both of these reactions are called a half-reaction.

In a redox reaction, the half-reaction Involving oxidation Is called oxidation half-reaction, and the half-reaction involving reduction is called reduction half-reaction.

Redox reaction = Oxidation half-reaction Reduction half-reaction

Reaction 1:

Reaction 2:

Reaction 3:

Oxidation State And Oxidation Number

According to electronic theory, redox reactions can be explained in terms of electron transfer. The electronic theory can be applied in the case of ionic compounds because, in the formation of ionic compounds, one reactant gives up the electron(s) and another reactant accepts those electron(s).

However, this theory cannot be applied in the case of a redox reaction involving covalent compounds due to the absence of cations or anions in such compounds. Thus this theory is unable to Identify the oxidant and reductant in such types of reactions.

To overcome this problem, the concept of oxidation number has been introduced. All types of redox reactions can be explained based on oxidation number. Each constituent element of any compound has a definite valency. Similarly, it can be assumed that each atom of any element has a definite oxidation number.

Oxidation state

An atom of an element converts into an ion when it loses or gains an electron. The loss of one or more electrons by an atom results in the formation of a cation, while the gain of one or more electrons produces an anion.

Cation is the oxidised state and anion is the reduced state of an element For example, the Na+ ion is the oxidised state of the Na atom and the Cl- ion represents the reduced state of the Cl -atom. The state of oxidation or reduction of an element presenting a compound is called the oxidation state of that element.

Oxidation state Definition:

The state of oxidation or reduction in which an atom of an element exists in a compound is called the oxidation state of the element in that compound.

From the element charge of the compound ion of an element present in an ionic compound, the oxidation state of that element can be easily determined. If the element exists as a cation in the compound, the element is said to be in the oxidised state. On the other hand, if the element exists as an anion in the compound, then the element is said to be in a reduced state.

Example:

In the formation of the compound ZnCl₂, the Zn atom loses two electrons to produce a Zn²⁺ ion and two CIatoms accept two electrons, one electron each to yield two Cl- ions. Zn²⁺ ion combines with two Cl- ions to form a ZnCl₂ molecule.

Thus, in ZnCl₂, the Zn -atom exists in the oxidised state, while the Cl -atom exists in the reduced state. Cations or anions derived from the same element may exist in different oxidation states in different compounds. For example, in CuCl₂, Cu exists as Cu²⁺, while in CuCl, it exists as Cu⁺ ion. Thus, the oxidation state of Cu in CuCl₂ is higher than that in CuCl.

Oxidation number

Oxidation number Definition:

The oxidation number of an element in a compound is a definite number, which indicates the extent of oxidation or reduction to convert an atom of the element from its free state to its bonded state in the compound.

If oxidation is necessary to effect such a change, then the oxidation number will be positive. The oxidation number will be negative when such a change requires reduction. The oxidation number of an atom or molecule in its free state is considered to be zero (0).

The number expressing the die oxidation state of the atom of an element in a compound denotes the oxidation number of the elements in the compound.

The oxidation number of elements in electrovalent compounds:

The charge that an atom of an element In the molecule of an electrovalent compound carries, is equal to the oxidation number of the element In a compound, According to the nature (positive or negative) of the charge, the oxidation number may be positive or negative. The number of electrons (s) lost or gained by an atom during the formation of an Ionic compound determines the oxidation number of the clement in that compound.

Examples:

ln NaCl, sodium and chlorine exist ns Na⁺ and (11- ions, respectively. So the oxidation numbers of sodium and chlorine are +1 and – 1, respectively. In FeCI₂> Iron and chlorine are present as Fe²⁺ and Cl^– Ions, so the oxidation numbers of iron and chlorine are +2 and -1, respectively.

The oxidation number of elements in covalent compounds:

The formation of a covalent compound does not involve the direct transfer of electron(s) between the participating atoms; instead, a covalent bond Is formed by the sharing of electrons.

When two atoms of different electronegativities form a covalent bond(s) through the sharing of one or more electron pairs, they do not get an equal share of the electrons.

The more electronegative atom acquires a greater possession of the shared electron pair (s) than the less electronegative atom.

As a result, the more electronegative atom acquires a partial negative charge. It Is assumed that the atom of the more electronegative element has gained electron(s) i.e. it has been reduced and the less electronegative atom has lost electron(s) i.e., It has been oxidised.

The oxidation number of an atom In a covalent compound is considered to be equal to the number of electron pair(s) the atom shares with one or more atoms of different electronegativities in the compound, If the atom concerned is of higher electronegativity, then Its oxidation number is taken as negative, and if It is of lower electronegativity, its oxidation number is taken as positive.

In the case of a molecule of an element, such as H₂, N₂, O₂, Cl₂ etc., the two atoms of the same electronegativity are covalently linked by sharing electron pair (s) between them. Hence, these two atoms use the electron pair(s) equally and none of the atoms acquire a positive or negative charge. Therefore, the oxidation number of the atoms In the molecule of these elements is regarded as zero.

Example:

1. In hydrogen chloride molecule (HCl) one electron pair exists between H and Cl-atoms, As the electronegativity of chlorine is more than that of hydrogen, the oxidation number of H^–  +1 and that of Cl =-1,

2. In water (H₂O) molecule, the O -atom shares two electron palms, one each with two separate H -atoms. As oxygen is more electronegative than hydrogen, the oxidation number of H -atom +l and that of O -atom =-2.

3. In a carbon dioxide molecule (CO₂), a carbon atom shares four electron pairs, to each with two separate O atoms. As oxygen has higher electronegativity than carbon, in CO₂ molecule, the oxidation number of C = + 4 and that of O =-2

4. In the ethylene (C₂H₄) molecule, two equivalent carbon atoms share two C=C electron pairs between themselves, Since these two electron pairs are equally shared by two C -atoms, they have no role in determining the oxidation number of C.

Again, each carbon atom shares two electron pairs with two separate H -atoms. As carbon is more electronegative than hydrogen, the oxidation number of each H -atom = + 1 and the oxidation number of each C -atom =-2.

Rules For Calculating the Oxidation Number Of An Element

The following rules are to be followed in determining the oxidation number of an element in a compound.

The oxidation number of an element in its free or elementary state is taken as zero (0).

Example:

⇒ \(\stackrel{0}{\mathrm{Z}} \mathrm{n}, \stackrel{0}{\mathrm{Cu}}, \stackrel{0}{\mathrm{C}} \mathrm{L}_2, \stackrel{\ominus}{\mathrm{P}}_4, \stackrel{0}{\mathrm{~S}}_8 \text {, etc. }\) Etc.,

The oxidation number of a monoatomic ion in an ionic compound is equal to its charge.

Example:

In FeCl₂, iron and chlorine exist as Fe²⁺ and Cl^–. So, in i.e., Cl₂, the oxidation number of Fe and Cl are +2 and -1 respectively.

In the case of a polyatomic ion, the sum of the oxidation numbers of all the atoms present in it is equal to the charge of the Ion.

Examples:

The sum of oxidation numbers of all the atoms presenting [Fe(CN)₈]4- =-1.0 The sum of the oxidation numbers of the atoms Cr₂O₇ ^2- ion =-2.

Determination of the oxidation numbers of the atoms in covalent compounds has been discussed in

The algebraic sum ofthe oxidation numbers of all atoms in a neutral molecule is zero(O).

Example: In the FeCl₃ molecule, the oxidation number of Fe is +3 and the oxidation number of Cl is -1. So, the total oxidation number of the atoms in the FeCl₃ molecule =(+3) + 3 × (- 1) = 0.

NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

The oxidation number of hydrogen:

In metallic hydride, it is always

In all hydrogen-containing compounds except for metallic hydrides, it is +1.

Example:

The oxidation number of oxygen in its compounds:

The oxidation number of oxygen in most compounds=-2

Example:

• In peroxide compounds (H₂O₂ Na₂O₂), the oxidation number of oxygen =-1.
• In superoxides (Example; KO₂), the oxidation number of oxygen =-1/2.
• Since fluorine is more electronegative than oxygen, the oxidation number of oxygen in F₂O = +2.
• In all fluorine-containing compounds, the oxidation number of fluorine =- 1.
• There are some elements which always show, fixed oxidation numbers in their compounds.

Example:

Alkali metals (Li, Na, K, Rb, Cs) always show a +1 oxidation state in their compounds. The oxidation number of alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra ) is +2 in their compounds. The oxidation number of Zn in its compounds is +2 and the oxidation number of Al in its compounds is +3.

The maxim oxidation1 number of an element cannot exceed its group numbering in the periodic table.

The following always have definite oxidation numbers in their compound

Calculation of oxidation number in some compounds

The oxidation number of an element in a compound can be calculated if the oxidation numbers of other elements in the compound are known.

The oxidation number of S in H₂SO₄:

Suppose, the oxidation number of S in H₂SO₄ = x.

The total oxidation number of two H-atoms in the H₂SO₄ molecule = 2 × (+1) = +2.

Total oxidation number of four O -atoms in H₂SO₄ molecule = 4 × (-2) =-8 .% Total oxidation number of all atoms in a H₂SO₄ molecule = +2 + x + (-8) = x- 6.

Now, the sum of the oxidation numbers of all atoms in a molecule = 0.

Therefore, x – 6 = 0 or, x = +6

∴ The oxidation number of S in H₂SO₄ = +6

The oxidation number of Cl in KClO₄:

If the oxidation number of Cl = x, then the total oxidation number of all the atoms in the KClO₄ molecule

=+1+x+ 4 × (-2) = x- 7

The sum of oxidation numbers for all the atoms present in a molecule = 0.

∴ x-7 = 0 or, x = +7

Hence, the oxidation number of Cl in KClO₄ =

The oxidation number of N in NH₄NO₃:

NH₄NO₃ is an ionic compound in which NH₄ and NO-₃ are the cation and anion, respectively.

Let the oxidation no. of in NH⁺₄ ion be x. Then the total oxidation no. of the atoms present in this ion = x + 4.

For a polyatomic ion, the oxidation no. of the ion is equal to its charge i.e., the oxidation number of the NH₄ ion

= +1. x + 4

= +1 or, x = -3.

Again, if the oxidation number of N in NO₃ is y, then y + 3 × (-2) = -1 or, y = +5.

Hence, in NH₄NO₃ the oxidation number of one Natom is -3 and that of another N -atom is +5.

The oxidation number of Cl in Ca(OCl)Cl:

In this compound. The Cl -atom in OCl^– is linked with the O-atom, and another Cl atom exists as the Cl^– ion. The oxidation number of the Cl -atom that exists as Cl^– ion =-1.

Let the oxidation number of the Cl atom in OCP be x.

∴ – 2+ x =-1 or, x =+l

So, in Ca(.OCDCl, the oxidation number of one Cl atom is 1 and that of another Cl^– atom is +1.

The oxidation number of Mn in KMnO₄: Suppose, the oxidation number ofMn.in KMn04 = x.

Total oxidation number ofthe atoms in KMnO₄ =+1 + x+ 4 × (-2) = x-7

∴ x-7 = 0 or, x = +7

Hence, the oxidation number of in KMnO₄ =+7

The oxidation number of P in H₄P₂O₇:

Let the oxidation number of P in H₄P₂O₇ be x. Hence, the total oxidation number of P-atom in the H₄P₂O₇ molecule

= 4 × (+1) -+2 × x + 7(-2)

= 4-+2x -14

= 2x- 10

2x- 10 = 0

Or, x = 5

Therefore, the oxidation number of P in H₄P₂O₇ = +5

Class 11 Chemistry Redox Reactions Notes

The oxidation number of Fe in Fe(CO)₅:

CO is a neutral ligand (molecule) and its oxidation number = 0. So, the oxidation number of Fein Fe(CO)- is zero (0).

The oxidation number of Fe in K₄[Fe(CN)₆]:

Suppose, the oxidation number of Fe in K4[Fe(CN)₆] =x. Hence, 4 × (+1) + x+ 6 × (-1) = 0

The oxidation number of an element in 3 compounds may be zero(0):

In compounds like C₆H₁₂O₆, HCHO, CH₂C₁₂ etc., the oxidation number of carbon is zero (0). Suppose, the oxidation number ofCin C₆H₁₂O₆ = x. So, 6x+ 12x(+1) + 6x(-2) = 0

∴ x = 0

Some exceptions regarding the determination of oxidation number

The anomaly fractional oxidation state:

Since electrons can not be transferred fractionally, the fractional oxidation state of an element seems to be a hypothetical case. But in compounds like Fe₃O_{4, and} NaS₄O6 the oxidation states of Fe, and S are \(+\frac{8}{3}\) and +2.5, respectively.

The fractional oxidation state is only the average oxidation state of an element when two or more of its atoms with different oxidation states are present in a compound. For such compounds, the actual oxidation state can be determined by knowing the structure of the compound.

1. The oxidation number of Cr in CrO₅:

According to the usual method, the oxidation number of Cr in the CrO₅ molecule would be +10.

However, the oxidation number of Cr can never exceed 6 because the total number of electrons in its 3d and 4s orbitals is 6. From the chemical structure of CrO₅, it can be shown that the oxidation number of Cr in CrO₅ is, in fact, +6.

Let the oxidation number of Cr in the CrO₅ molecule be x.

x +  1 × (-2) (for O )+ 4 × (-1)(for O-atoms linked in O – O bond ) = 0

x = +6.

Hence, the oxidation number of Cr in CrO₅ = +6.

2. The oxidation number of S in H₂SO₅:

According to the tyre .usual method, the oxidation number of sulphur in the H₂SO₆ molecule is +8

The oxidation number of sulphur can never exceed + 6. The chemical structure of H₉SO₆ shows that the oxidation number of sulphur in H₂SO₅ is

Suppose, the oxidation number of S in H₂SO₅ = x

2 × (+1)( For H-atoms) + x + 2 × (-1) ( (For O-atoms held by O—O bond + 3 × (-2) ( For other- O-atoms)

x = + 6

Hence, the oxidation number of the S -atom in H₂SO₅ = +6

3. The oxidation number of S in Na₂S₂O₃:

According to the usual method, the average value of oxidation numbers of S in Na₂S₂O₃ molecule = +2

However the reaction of Na₂S₂O₃ with dilute H₂SO₄, one of the two S -atoms in Na₂S₂O₃ is precipitated as sulphur and the other is oxidised to SO₂.

Therefore, the two S -atoms in the Na₂S₂O₃ molecule are not identical. Consequently, their oxidation numbers cannot be the same. The chemical structure of the Na₂S₂O₃ molecule indicates that the two sulphur atoms in it are linked by a coordinate bond.

The oxidation number of the S -atom accepting the electron pair in the coordinate bond is considered to have an oxidation number of -2. If the oxidation number ofthe other S -atom is taken as .x, then.

2 × (+1)( For Na-atoms)  + 3 × (-2) ( (For O-atoms) + x × 1 + 1 × (-2) = 0 ( (For S-atom by coordinate bond)

∴ x = +6

Therefore, in the Na₂S₂O₃ molecule, the oxidation number of one S -atom is -2 and that of the other is +6

Redox Reactions Chapter 8 NCERT Notes

4. The oxidation number of S in Na₂S₄O₆:

According to the usual method, the average value of oxidation numbers of S in Na₂S₄O₆ molecule would be + 2.5

In atoms, this molecule is covalently oxidationlinkednumberis zero. If of the two oxidation sulphur numbers of each ofthe remaining two S -atoms is x, the

x × 2 (For S)+ 2× 0 (For S-S)+ 6 ×(-2)(For O) ×  2×  (+1) (For Na)= 0

Or 2x- 12 +12 = 0, x =+5

Therefore, the oxidation number of each of the two remaining S -atoms in Na₂S₄O₆ is +5

Explanation Of Oxidation-Reduction In Terms Of Oxidation Number

According to the concept of oxidation number, oxidation is a chemical reaction in which the oxidation number of atoms increases and reduction is a chemical reaction in which the oxidation number of an atom decreases.

So, oxidation means an increase in oxidation number, whereas reduction means a decrease in oxidation number. Examples of oxidation and reduction are as follows.

Explanation of oxidation-reduction reaction

Consider the reaction of HNO₃ with H₂S, forming nitric oxide (NO) and sulphur. In this reaction, the oxidation number of N decreases from +5 in HNO₂ to +2 in NO and the oxidation number of S increases from -2 in H₂S to 0 in S. So, the reaction brings about a reduction of HN03 and oxidation of H₂S.

In a redox reaction, all the atoms of a participating reactant do not change their oxidation number. Only one atom of the reactant changes oxidation number. This element is called an effective or reactive element.

In the previous example, only the N -atom of HNO₃ changes oxidation number. The oxidation numbers of hydrogen or oxygen remain the same before and after the reaction.

The reaction of FeSO₄ with KMnO4 acidified with dilute H₂SO₄ results in K₂SO₄, MnSO₄, FeSO₄ and H₂O.

In this reaction, the oxidation number of Mn decreases (+ 7→ + 2) and the oxidation number of Fe increases (+2 → + 3). Therefore, in the reaction, KMnO₄ is reduced and FeSO₄ is oxidised

CBSE Class 11 Redox Reactions Chapter 8 Notes

Identification of oxidant and reductant based on oxidation number

In redox reactions, the substance that gets oxidised is the reducing agent and the substance that gets reduced is the oxidising agent. Based on oxidation number, it can be stated that in a redox reaction, the substance in which the oxidation number of an atom increases is the producing agent and the substance in which the oxidation number of an atom decreases is the oxidising agent.

Example:

In presence of H₂SO₄, K₂Cr₂O₇ reacts with KI to form I₂ and chromic sulphate [Cr₂(SO₄)₃]

Here, the oxidation number of Cr decreases from +6 to +3 and the oxidation number of 1- increases from -1 to 0. Therefore, K₂Cr₂O₇ acts as an oxidising agent and KI acts as a reducing agent.

How a redox reaction is identified:

At first oxidation number of each of the constituent elements of the participating substances is assigned. If the oxidation numbers of the elements change, then the reaction is identified as a redox reaction. If none of the elements shows any change in oxidation number, then the reaction is not a redox reaction.

Example: Identify whether the given two reactions are redox reactions or not

In this reaction, the oxidation numbers of all the atoms of the participating substances remain the same. Hence, it is not a redox reaction.

In this reaction, the oxidation number of N increases 0) and the oxidation number of oxygen decreases (0 →2). Hence, it is a redox reaction.

Auto Oxidation-Reduction Reactions

There are some redox reactions in which the same substance gets partially oxidised and reduced. This type of reaction is termed an auto oxidation-reduction reaction.

Example: Potassium chlorate (KCLO₃) on heating decomposes to produce KCl and O₂ gas:

In this reaction, the oxidation number of Cl decreases from +5 to -1 and the oxidation number of oxygen increases from- 2 to 0. So, in this reaction of KClO₃, one atom (O) is oxidised and the other (Cl) is reduced.

Lead nitrate undergoes thermal decomposition to produce PbO, NO gas and 02 gas:

In this reaction, the oxidation number of the N -atom reduces from +5 to +4 and the oxidation number of the O -atom increases from -2 to zero (0 ). Hence, in the reaction, the N atom is reduced and the O -atom is oxidised. So, Pb(NO₃)₃ in its thermal decomposition undergoes oxidation and reduction simultaneously.

Ammonium nitrate (NH₄NO₃) on heating decomposes to produce water vapour, N₂ gas and O₂ gas

NFI₄NO₃ is an ionic compound, consisting of ammonium cation (NH⁺₄) and nitrate anion (NO₃). The oxidation numbers of the N -atom in NH₄⁺  and NO₂ ions are -3 and 5 respectively.

In this reaction, the oxidation number of N in the NH₄ ion increases (-3 → 0) and the oxidation number of N in NO₃ decreases (+ 5 → 0). Therefore, NH₄NO₃ undergoes oxidation and t In this reaction, reduction at die same time.

Disproportionation And Comproportionation Reactions

Disproportionation reaction in which an element of a reactant undergoes oxidation and reduction simultaneously, resulting in two substances in which one of the elements exists in a higher oxidation state and the other exists in a lower oxidation state

Examples:

In the reaction of chlorine with cold and dilute NaOH solution, sodium hypochlorite (NaOCl) and sodium chloride are formed. In this reaction oxidation number of chlorine decreases (0→ -1) and increases (0→ +1) at the same time. thus, chlorine is simultaneously oxidised and reduced in the reaction.

Therefore, this reaction is an example of a disproportionation reaction.

When white phosphorus is heated with a caustic soda solution, phosphine (PH₄) and sodium hypophosphite (NaH₂PO₂) and produced.

Here P. undergoes oxidation and reduction simultaneously. In one of the products (PH₄), P exists in a lower oxidation state (-3) and the other product (NaH₄PO₂), it exists in a higher oxidation state (+1). Hence, the tills reaction is an example of a disproportionation reaction.

NCERT Class 11 Chemistry Redox Reactions Chapter 8

Comproportionation Reaction:

It is a reaction in which two reactants containing a particular element but in two different oxidation states react with each other to produce a substance in which the said element exists in an intermediate oxidation state.

Therefore, a comproportionation reaction is the opposite of a disproportionation reaction.

Example: KBrO₃ reacts with KBr in an acidic medium to produce Br₂.

In this reaction, the oxidation number of one Br-torn decreases (from +5 to. 0) and that of another Br-atom increases (from -1 to 0 ). Br₂ is formed by the oxidation of KBrO₃ and the reduction of KBr. The oxidation number of Br₂ is zero, which is intermediate between the oxidation numbers of Br atoms in KBrO, (+5 ) and KBr (-1 ). Therefore, it is a comproportionation reaction.

Equivalent Mass Of Oxidant And Reductant

The equivalent mass of oxidants and reductants is calculated by following two different methods. In one method, the calculation is done in terms of several electron (s) gained by an oxidant or the number of electrons (s) lost by a reductant. The other method takes into account the change in the oxidation number of an element present in the oxidising and reducing agents.

Oxidation number method:

The equivalent mass of an oxidant or reductant denotes the number obtained by dividing the molecular mass of the oxidant or reductant with the change in oxidation number of an element in the oxidant or reductant in their respective reduction or oxidation reaction.

⇒ \(\begin{aligned}
& \text { Equivalent mass of the oxidant }
\end{aligned}=\frac{\text { Molecular or formula mass of oxidant }}{\begin{array}{c}
\text { Total change in oxidation number } \\
\text { of an element present in a molecule } \\
\text { of the oxidant during its reduction }
\end{array}}\)

⇒ \(\begin{aligned}
& \text { Equivalent mass of the reductant }
\end{aligned}=\frac{\text { Molecular or formula mass of reductant }}{\begin{array}{c}
\text { Total change in oxidation number } \\
\text { of an element present in a molecule } \\
\text { of the reductant during its reduction }
\end{array}}\)

Determination of equivalent mass of oxidants 

Determination of equivalent mass of reductants

Electronic method: Equivalent mass of an oxidant or reductant is a number obtained by dividing the molecuslar mass or formula mass of an oxidant or reductant by the number ofelectron(s) gained or lost by a molecule of that oxidant or reductant during reduction or oxidation ofthe respective compound.

The equivalent mass of an oxidant or a reductant is formulated  as:

Equivalent mass ofthe oxidant = \(\frac{\text { Molecular or formula mass of oxidant }}{\text { Number of electron(s) gained by each molecule of oxidant during reduction }}\)

Equivalent mass bf the reductant = \(\frac{\text { Molecular mass or formula mass of a reductant }}{\text { Number of electron(s) lost by each molecule of reductant during oxidation }}\)

Class 11 Chemistry Redox Reactions Example Solutions

Determination of equivalent mass of oxidants

Determination of equivalent mass of reductants

NCERT Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics Short Question And Answers

Question 1. What is an adiabatic system? Is this an isolated system?
Answer:

An adiabatic system Is a closed system that can exchange several forms of energy (for example Work) but not heat with its surroundings. It is not an isolated system as an isolated system cannot exchange either matter or energy with its surroundings

Question 2. A closed container with impermeable diathermal walls contains some amount of gas. If the gas is considered to be a system, what type of system will it be X is a state function of a thermodynamic system. How are its infinite and infinitesimal changes denoted?
Answer:

As the walls of the container are impermeable, flow of matter into or out ofthe system is not possible. Again, the walls are diathermal. Thus, the exchange of heat between the gas and the surroundings is possible. Hence, the gas is a closed system.

Question 3. A closed system participates in the following process: A→B→C. In step A→B heat absorbed by the system = q cal and in step B→C, heat released by the system = qcal. Therefore, in this process, the sum of the heat absorbed and heat released by the system is zero. Is this an adiabatic process? Give reason.
Answer:

In an adiabatic process, the system does not exchange heat with its surroundings at any step in the process. In the given process, the system absorbs heat in one step and releases heat in the other step. So, this process cannot be regarded as an adiabatic process.

NCERT Class 11 Chemistry Chapter 6 Short Question and Answers

Question 4. One mole of an ideal gas participates in a cyclic reversible process as described. Indicate the type of processes the system undergoes in the steps AB, BC and CA. Assume T₂>T₁.
Answer: AB:

It is an isochoric process as the volume of the system remains unaltered in this step. BC: It is given that T₁<T₂– Again, the given indicates V₂> V₁. This means that the volume of the system increases with a decrease in temperature. This happens in case of an adiabatic expansion of a gas. Therefore, the BC step indicates an adiabatic process.

Question 5. Calculate the work done in the following process which an
ideal gas undergoes
Answer:

⇒ \(\text { 1st step: } w_1=-n R T_1 \ln \frac{V_2}{V_1}\)

⇒  \( 2 \text { nd step: } w_2=-n R T_1 \ln \frac{V_1}{V_2}\)

So, total work \(w_1+w_2=-n R T_1 \ln \frac{V_2}{V_1}-n R T_1 \ln \frac{V_1}{V_2}\)

⇒ \(=-n R T \ln \frac{V_2}{V_1}+n R T \ln \frac{V_2}{V_1}=0\)

Question 6. In which of the following reactions is the work done zero? Assign the sign of w (+ ve or- ve)for the cases in which work is involved.
Answer:

In a chemical reaction, pressure-volume work, ω = -P_exΔV = -ΔnRT; where Δn = total number of moles of gaseous products – total number of moles of gaseous reactants.

In reaction 3 Δn = 0Δ, So, w =0

In reaction 1 Δn = 2-(1+2)=-1

So, w = -ΔnRT = RT, i.e., w> 0

In reaction Δn=l. So, ω= -ΔnRT =-RT, i.e., w < 0.

Question 7. Among the following processes identify those In which the change in internal energy (Δ U) Is zero: Isothermal compression of ideal gas Adiabatic expansion of ideal gas Free adiabatic expansion of an ideal gas Reversible cyclic process Irreversible cyclic process.
Answer:

The change in internal energy of an ideal gas in its isothermal compression is zero, When an ideal gas undergoes an adiabatic expansion, its internal energy decreases. In the adiabatic free expansion of an ideal gas, the internal energy of the gas remains the same, Since U is a state function, its change in any cyclic process (reversible or irreversible) will be zero.

Question 8. Write down the mathematical form of the first law of thermodynamics for an infinitesimal change that involves only pressure-volume work. Write down the form of this equation if the above change occurs reversibly.
Answer:

In case of an infinitesimal change, the mathematical form of the first law of thermodynamics is: dU= δq + δw; where δq = heat absorbed by the system, δw = work done on the system and dU is the change In Internal energy of tyre system.

For an infinitesimal change involving only P-V work, δw=-P_exdV. So, for an infinitesimal change Involving only P-V work, the form of the first law of thermodynamics will be, Δw=-P_exdV

Question 9. A closed system undergoes a process A→B. If it occurs reversibly, then the system absorbs qy amount of heat and performs a amount of work. However, if it occurs irreversibly, then the system absorbs the q₂ amount of heal and does the w₂ amount of work. Is (q₂ + w₁) greater than, less than or equal to (q₂+ w₂)?
Answer:

For a reversible process:

ΔU₁ = q₁ + w₁ and for the irreversible process: ΔU₂ = q₂+w₂.

In both processes, the initial state (A) and final state (B) of the system are identical. Since U is a state function, its change in a process depends only on the initial and final states ofthe system, and not on the nature of the process.

As the initial and final states in both processes are identical, the change in internal energy in both cases will be the same. Therefore, ΔU₁ = ΔU₂ and q₁+ ω₁ = q₂ + w₂

Question 10. C_p-C_y = x J-g^-1.K^-1 and C_p-C_y = x J-g^-1.K^-1 ] mol^-1. K^-1 for an ideal gas. The molecular mass of the gas is M then establishes a relation among x, X and M.
Answer:

For a substance, molar heat capacity = specific heat capacity x molar mass.

Therefore, \(C_{P, m}=C_P \times M \text { and } C_{V, m}=C_V \times M\)

Given: Cp,m- Cv,m = XJ mol^-1. K^-1

∴ CpxM-CvxM = X

or, (Cp-CV)M = X; hence, X = Mx

Question 11. Why is the sign of ΔH negative for an exothermic reaction and why is it positive for an endothermic reaction?
Answer:

In a chemical reaction, the change in enthalpy, AH = sum of the total enthalpies of products – Sum of the total enthalpies of-reactants

= \(\Sigma H_P-\Sigma H_R\)

In case of an exothermic reaction

⇒  \(\Sigma H_P-\Sigma H_R\), and hence ΔH<0;

While for an endothermic reaction \(\Sigma H_P-\Sigma H_R\) resulting ΔH>0.

Question 12. Mention the standard stales of the following elements at 25°C and later: carbon, bromine, Iodine, sulphur, oxygen, calcium, chlorine, fluorine and nitrogen.
Answer: At 25C; and 1 atm, the standard states of the given elements are—

1. Carbon: C(s, graphite);
2. Bromine: Br₂(Z);
3. Iodine: l₂(s):
4. Sulphur: S(s, rhombic);
5. Oxygen: O₂(g);
6. Calcium: Ca(s)
7. Chlorine: Cl₂(g);
8. Pluorine: P₂(g) ;
9. Nitrogen: N₂(g)

Question 13. Give an example of a physical change for each of the following relations between
ΔH And ΔU:

1. ΔH > ΔU
2. ΔH < ΔU
3.  ΔH≈ ΔU

Answer:

The equation ΔH= ΔH + ΔnRT can be used in case of a process involving phase change ofa substance.

1.  \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2 \mathrm{O}(g) \text {. Here, } \Delta n=+1 \text {. So, } \Delta H>\Delta U \text {. }\)

2. \(\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) \text {. Here, } \Delta n=-1 \text {. So, } \Delta H<\Delta U \text {. }\)

Chemical Thermodynamics Class 11 Short Questions

Question 14. Which element in each of the following pairs has the standard heat of formation to zero?

1. [O₂(g), O₃(g)] 
2. [Cl₂(g), Cl(g) ]
3. [S (s, rhombic), S (s, monoclinic)]?

Answer:

The enthalpy of formation of an element in its standard state is zero, at 25 the standard state of oxygen, chlorine and sulphur are

O₂(g)> Cl₂(g) and S (s, rhombic). So, at 25 °C, the standard heats of formation of O₂(g), CI₂(g) and S(s, rhombic) will be zero.

Question 15. “At 25°C the standard heat offormation ofliquid benzene is + 49.0 kj.mol^-1 What does it mean
Answer:

At 25°C, the standard enthalpy of formation of liquid benzene is +49.0 kj.mol^-1. This means that at 25 °C and 1 atm when 1 mol liquid benzene forms from its constituent elements, the enthalpy change that occurs is +49.0 kj. In other words, at 25 °C and atm pressure, the change in enthalpy in the following reaction is +49.0 kj

6C(s, graphite) + 3H₂(g)→C₆H₆(l)

Question 16. The standard heat of sublimation of sodium metal is 108.4 kj.mol^-1. What is its standard heat of atomization?
Answer:

At 25 °C, the standard state of sodium is Na(s). The sublimation process of Na(s) is Na(s)→Na(g). At 25 °C the enthalpy change process is equal to the sublimation enthalpy of Na-ihetal. Again, in the above process, I mol of Na(g) is formed from Na(s). So, at 25°C, the enthalpy change in this process is equal to the standard enthalpy of atomisation of sodium. Thus, the standard enthalpy of atomisation of sodium is 108.4 kj. mol^-1

Question 17. What will be the sign of ASsys (+ve or -ve ) in the process of—

1. The vaporisation of a liquid
2. Condensation of a vapour
3. Sublimation of a solid.

Answer:

In the vaporisation of a liquid (liquid → vapour) \(\Delta S_{\text {system }} \text { is +ve. }\) This is because a substance in its vapour state possesses greater entropy than its liquid state.

Question 18. Consider the reaction, A → 2B, if the free energy per mole of A is GA and that of B is GB then what will the relation be between GA and GB when reaction 1 occurs spontaneously and 2 is at equilibrium?
Answer:

For a spontaneous reaction at a given temperature and pressure, ΔG < 0. Therefore, 2G_B< G_A

At a given temperature and pressure, for a reaction at equilibrium, AG = 0

Therefore, ΔG = 2G_B-G_A = 0 or, G_A = 2G_B

Question 19. “The amount of heat present in hot water is greater than that in cold water”—explain whether the statement is correct or not.
Answer:

1. The statement is wrong because heat can never be stored in any system as it is a form of energy In transit.
2. During a process beat appears at the boundary of a system. Heat does not exist before and after the process.

Question 20. Give examples of two processes involving only P-V work, where the system does not perform any work.
Answer:

1. During the expansion of a gas against zero external pressure, work done is zero.
2. If a process, involving only pressure-volume work, is carried out at constant volume, then work done in the process will be zero. For example, in the case of the vaporization of water in a closed container of fixed volume, the work done is zero

Question 21. A plant is growing. What do you think of the entropy changes of the plant and its surroundings?
Answer:

The entropy decreases during the growth of the plant (i.e., system) because the ordered structure of the plant is formed during 1(8 growth. However, the entropy of the surroundings increases during the process. The increase in entropy of the surroundings is much greater than the decrease in entropy of the system. As a result, the net entropy change of the system and its surroundings is always positive during the growth of a plant.

Question 22. When does an adiabatic process become isentropic?
Answer:

1. In a process, if the entropy of a system remains unchanged, then the process is called isoentropic.
2. In a reversible adiabatic process \(\delta q_{r e v}\) =0 So the entropy change \(d S=\frac{\delta q_{r e \nu}}{T}=0\) Therefore, a reversible adiabatic process is isentropic.

Question 23. mol of an ideal gas is freely expanded at a constant temperature. In this process, which of the quantities or quantities among w, q, AU, and AH are 0 or >0 or <0?
Answer:

1. During the isothermal free expansion of a gas, the work done is zero. So w = 0. Again internal energy and enthalpy remain the same during the isothermal expansion of an ideal gas. Therefore, ΔU = 0 and ΔH = 0.
2. According to the first law of thermodynamics, ΔU = q + w. For the given process ΔH = 0 and w = 0.
3. Therefore, q = 0. Thus, for isothermal free expansion of lmol of an illegal gas q = 0, w = 0, ΔU = 0, ΔH = 0.

Question 24. In process A →B →C → D, the heat absorbed by the system in steps A → B and B C are q₁ and q₂, respectively, and the heat released by the system in step C→ D is q₃. If q₁ + q₂ + q₃ = o, then will the process be adiabatic?
Answer:

In an adiabatic process, heat is not exchanged between a system and its surroundings at any stage of the process. In the given process, heat is being exchanged between the system and the surroundings. Thus, the process is not adiabatic though the sum of the amounts of heat absorbed and released is zero for the process.

Question 25. Why is infinite time required for the completion of an ideal reversible process?
Answer:

In an ideal reversible process, the system maintains equilibrium at every intermediate step and the process is extremely slow. Thus, from a theoretical point of view, an ideal reversible process should require an infinite time for Its completion.

Question 26. At 25°C, is the standard reaction enthalpy for the reaction 2H(g) + O(g) → H₂O(I) the same as the standard enthalpy of formation of H₂O(f)
Answer:

1. The given reaction does not indicate the formation reaction of H₂O(f) because the standard states of hydrogen and oxygen at 25°C are H₂(g) and O₂(g), respectively.
2. Hence, the standard reaction enthalpy of the given reaction is not the same as the standard enthalpy of the formation of H₂O(l).

NCERT Solutions Class 11 Chemistry Chapter 6 Short Q&A

Question 27. Which condition does not satisfy the spontaneity criteria of a reaction at constant temperature and pressure: ΔH<0, ΔS<0, ΔH>0, ΔS<0, ΔH>0, ΔH<0, ΔS>0?
Answer:

1. At constant temperature and pressure, a reaction will be spontaneous if ΔG = -vc for the reaction at that temperature and pressure.
2. If ΔG = +vc, the reaction will be nonspontaneous.
3. At a constant temperature and pressure, ΔG = ΔH- TΔS. If ΔH > 0 and ΔS<0, then ΔG = +i/e.
4. Thus, a reaction will be noil-spontaneous if ΔH > 0 and ΔS < 0.

Question 28. According to the definition of a thermodynamic system, which system do living beings belong to, and why?
Answer:

According to the definition of a thermodynamic system, every living being in nature belongs to an open system.

Explanation: All living beings (systems) take food (matter) from the surroundings and excrete waste materials (matter) to the surroundings. They also exchange heat (energy) with the surroundings.

Question 29. Comment on (lie thermodynamic stability of NO(g).
Given:

⇒ \(\frac{1}{2} \mathrm{~N}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{NO}(g); \Delta_r H^0=90 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒  \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{NO}_2(g); \Delta_r H^0=-74 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)
Answer:

1. For the first reaction, refer to the standard enthalpy of formation (ΔrH°) for NO because 1 mol of NO forms from its constituent elements.
2. The positive value of Δr-Hº of a compound implies that the compound has enthalpy (or energy) than its constituent elements. Hence, the compound will be unstable.
3. Therefore, the positive value of ΔH° for the first reaction indicates that NO is unstable.

Question 30. Calculate the entropy change in the surroundings when 1.00 mol of ΔfHº(J) is formed under standard conditions. AH° = -286 kj.mol^-1
Answer:

For the given process

⇒ \(\Delta S_{\text {surr }}=-\frac{\Delta_f H^0}{T}=-\frac{-286 \times 10^3 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{298 \mathrm{~K}}\)

= 959.73 J-K^-1– mol^-1

Question 31. For the reaction, 2A + B→C; ΔH = 400 kj. mol 1 & ΔS = 0.2 kj.K^-1 .mol^-1 at 298 K. At what temperature will the reaction become spontaneous considering ΔH, ΔS to be constant over the temperature range?
Answer:

We know, ΔG = ΔH- TΔS. For a spontaneous reaction at a given temperature and pressure ΔG < 0.

Given:

ΔH = 400 kj.mol^-1 and

ΔS = 0.2 kj. K^-1 mol^-1

So, ΔG = (400- T ×  0.2) kj. mol^-1

According to this relation, ΔG will be <0 when T × 0.2 > 400 i.e., T> 2000K.

Question 32. For the reaction 2Cl(g)→Cl₂(g), what are the signs of ΔH andΔS?
Answer:

The process involves the formation of a bond, which is always exothermic. Hence, AH < 0 for this process. The no. of gaseous particles decreases in the process. Consequently, the randomness of the system decreases. Hence, ΔS < 0 for this process.

Question 33. State the second law of thermodynamics based on entropy. The boiling point of ethanol is 78.4°C. The change in enthalpy during the vaporization of ethanol is 96 J- mol^-1. Calculate the change in entropy of vaporization of ethanol.
Answer:

We know,

⇒ \(\Delta S_{\text {vap }}\)

= \(\frac{\Delta H_{\text {vap }}}{T_b}\)

= \(\frac{96}{(273,+78.4)} \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)

= \(0.2732 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)

Question 34. For the following reaction at 298 K 2X + Y → Z , ΔH = 300 kj.mol^-1 and ΔS = 0.2 kj K^-1.mol^-1 At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
Answer:

For spontaneous process, ΔG < 0

∴ \(\Delta H-T \Delta S<0 \text { or, } \Delta H<T \Delta S_1 \text { or, } \frac{\Delta H}{\Delta S}<T\)

⇒ \(\text { or, } \frac{300}{0.2}<T \text { or, } 1500<T\)

Therefore, the given reaction becomes spontaneous above 1500 K temperature.

Question 35. What is meant by an isolated system?
Answer:

We know, ΔG = ΔH- TΔS

or, AG = (29.3- 298 ×104.1 ×10^-3) kj.mol^-1

= -1.7218 kj-mol^-1

At a particular pressure and 298 K temperature, the free energy change of the given reaction is negative which indicates the spontaneity of the reaction.

Question 36. The initial pressure, temperature & volume of 1 mol of gas are P1, T₁ and V₁ respectively. The state of the gas is changed in the following two ways. Will the internal energy change be the same in both cases?
Answer:

The internal energy of a system is a state function. The change in internal energy in a process depends only on the initial and final states of the system. It does not depend on the path used to arrive at the state. Since the initial and final states are the same in 1 and 2, the internal energy change will also be the same.

Question 37. If one mole of an ideal gas is expanded in the following two ways, then will the value of P₂ and P₂ be greater than, less than or equal to P₁?
Answer:

The temperature of the gas remains the same during isothermal expansion. Therefore, P₂ < P₁. On the other hand, the temperature of the gas decreases during an adiabatic expansion. Therefore, in the process, T₂ is less than T₁. In this process. P₁ < P₂ since T₂<T₁

Question 38. Why does the value of All for a chemical reaction depend on the physical states of the reactant (s) and produces)?
Answer:

In the case of solids and liquids, ΔH = ΔH for a chemical reaction (as AV is negligible here). If the participating substances are gases, then ΔH = ΔH + ΔnRT. Hence, AH for a chemical reaction depends on the physical states of the reactant(s) and product(s)

Question 39. Will the transformation ofice into water be spontaneous -2°C and latm pressure? Will the reverse process be spontaneous at this pressure and temperature?
Answer:

No. The transformation of ice into water at -2°C and 1 atm pressure is not spontaneous. This is because the sum ofthe increase in entropy of the system and the decrease in entropy of the surrounding is less than zero.

The reverse process, i.e., the transformation of water into ice is spontaneous. This is because at -2°C and 1 atm pressure the sum of die decreases in the entropy ofthe system and the increase in entropy of the surroundings is greater than zero.

Class 11 Chemistry Chapter 6 Chemical Thermodynamics Short Answers

Question 40. Heat is not exchanged between the system and its surroundings during the free expansion of an ideal gas. Therefore, in this process, q = 0. Will the change in entropy in this process be zero?
Answer:

1. In the free expansion of an ideal gas, no exchange of heat takes place between the system and its surroundings.
2. Because of the expansion, the volume of the gas increases, and the larger space is now available to the gas molecules for their movement.
3. This results in an increase in randomness in the system, and hence the entropy ofthe system increases.

Question 41. Why does the entropy of the gaseous system Increase with the temperature rise?
Answer:

1. Due to large intramolecular distance and weak intermolecular forces, the molecules in a gas can move about freely.
2. The motion ofthe molecules becomes more random and disordered with the rise in temperature as the average speed of the molecules increases.
3. Now, the entropy ofa system is a measure of the disorderliness of the constituent molecules. Therefore, the entropy ofa gas increases with the temperature rise.

Question 42. Give an example of a process for each of the given cases: ΔG = 0, ΔS <0, ΔG= 0, ΔS > 0 ΔG < 0, ΔS > 0 ΔG<0, ΔS<0 in a system.
Answer:

Fusion of ice at 0°C and 1 atm pressure.

Condensation of water vapour at 100°C and 1 atm.

⇒ \(\mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \text { at } 25^{\circ} \mathrm{C} \text { and } 1 \mathrm{~atm}\)

⇒ \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3(\mathrm{~g}) \text { at } 25^{\circ} \mathrm{C} \text { and } 1 \mathrm{~atm} .\)

Question 43. A particular amount of an ideal gas participates in a reversible process as given in the figure. What type of process is this? Explain the changes in each step.
Answer:

1. This process is cyclic because the system returns to its initial state after undergoing consecutive processes AB, BC and CA.
2. In step AB, the gas expands reversibly at constant pressure. p A Hence, step AB indicates an isobaric change. In step BC, the temperature of the gas decreases at constant volume.
3. Thus BC indicates an isochoric change.
4. In step CA, the gas is compressed reversibly at a constant temperature. Thus CA indicates a reversible isothermal change.

Question 44. The transformation of A to B can be carried out in the following two ways in which the initial and final states are identical.

What will be the value of ΔH during the transformation of C to B?

Answer:

According to Hess’s law, the change in enthalpy in the process

1 = total change in enthalpy in process  2

∴ -xkJ =-yKJ + ΔH = (y-x)kJ.

Question 45. Write three differences between reversible and irreversible processes. Melting office at 0°C and 1 atm pressure is a reversible process— explain.
Answer:

1. It is a reversible process. Ice melts at 0°C under normal atmospheric pressure. Latent heat for the fusion of Ice Is lit) cal .g-1, i.e., 80 cal of heat is required to melt logfile. If 80 cal heat IB is extracted from the surroundings, 1 g of ice gets converted Into water.
2. Therefore, at normal pressure and temperature, ice and water remain in an equilibrium state.
3. By Increasing or decreasing the value of the driving force (by the supply or extraction of heat) the process can be made to move In the forward or backward direction.
4. So, the melting of ice at normal atmospheric pressure and temperature is an example of a reversible process.

Question 46. The boiling point of benzene is 80.1 °C. At ordinary pressure and 70°C, the benzene vapour spontaneously transforms into liquid benzene. In this process, what will the signs of ΔH, ΔS and ΔG be?
Answer:

1. The entropy of the system decreases when a vapour transforms into a liquid.
2. So ΔS < 0. Again, the condensation is an exothermic process. So, in this process, ΔH < 0.
3. Under the given conditions, the benzene vapour spontaneously condenses into liquid. So, in this process ΔG < 0.

Question 47. What is meant by the terms change of entropy (ΔS) and change in free energy (ΔG) of a system? Write down the mathematical relation between them. At 0°C, liquid water and ice remain in equilibrium. If lg of liquid water under equilibrium conditions is converted to ice, explain with reason whether the process is endothermic or exothermic.
Answer:

1. In the conversion of water into ice, the entropy of the system decreases, and Hence AS_sys < 0
2. We know, ΔG = ΔH- TΔS
3. For the given process, ΔG< 0. As the process occurs spontaneously, ΔG < 0 for the process.
4. According to the relation (1), if ΔS<0, then AG will be negative only when ΔH < 0. So, the process is exothermic.

Question 48. Given: C(s) + O₂(g)→CO₂(g); ΔH = -393.5 kj 2H₂(g) + O₂(g)→2H₂O(g) ; ΔH = -571.6 kj. Calculate ΔH of the reaction:

⇒ \(\mathrm{C}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2(\mathrm{~g})\)

Answer:

C(s) +O₂(g)→CO₂(g); ΔH = -393.5 kj……………… (1)

2H₂(g) + O₂(g)→2H₂O(g) ; ΔH = -571.6 kJ……………… (2)

Subtracting equation (2) from equation (1), we have C(s) + 2H₂O(g)→CO₂(g) + 2H₂(g);

ΔH = [-393.5- (-571.6)] k] = 178.1J.

NCERT Class 11 Chemistry Chapter 6 Chemical Thermodynamics Solutions

Question 49. Calculate the enthalpy of the formation of liquid ethyl alcohol from the following data.
Answer:

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H=-1368 \mathrm{~kJ}\)

⇒ \(\mathrm{C}(s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H=-393 \mathrm{~kJ}\)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \( \Delta H=-287 \mathrm{~kJ}\)

Question 50. N₂(g) + 3H₂(g)Δ₂NH₃(g) ; ΔrH°=-92.4 kj.mol-1. What is the standard enthalpy of the formation of NH₃?
Answer:

⇒ \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g}) ; \Delta_r H^0=-92.4 \mathrm{~kJ}\)

⇒ \(\frac{1}{2} \mathrm{~N}_2(g)+\frac{3}{2} \mathrm{H}_2(g) \rightarrow \mathrm{NH}_3(g) ; \Delta_r H^0=-46.2 \mathrm{~kJ}\)

Equation (1) represents the formation of NH₃(g) from the constituent elements. So, the standard enthalpy change for the reaction represented by equation (1) = the standard enthalpy of formation for NH₃(g) = -46.2 kj.mol^-1

Question 51. For the reaction, 2A(g) + B(g)→2D(g) ; ΔUº=-l0.5kJ and ASº = -44.11.K^-1 Calculate ΔG° for the reaction and predict whether it may occur spontaneously.
Answer:

The temperature has not been mentioned in the problem. Here, the calculation has been done by considering temperature as the normal temperature (298 K). For the reaction, An = 2- (2 + 1) = -1 . So, for this reaction,

ΔH° =ΔU° + ΔnRT =- 10.5-1 × 8.314 × 10^-3×298 kj.

=-12.98kj

We know, ΔG° = ΔHº -TΔSº

∴ AGº =(- 12.98 + 298 × 44.1 ×10^-3) kl =0.16kl

The positive value. of AG° indicates; that the cannot occur spontaneously.

Question 52. The equilibrium constant for a mission is 10. Find the value of AG0 t R a 0.814 MC ^–1mol^–1, T = 300K.
Answer:

We know, ΔG° w -2.303/log K

Given K=10 and T = 300k

∴ A (1° a 11.303 × 11.3 1 4 × 300 log10).mol^-1

=-5.74kJ.mol^-1

NCERT Class 11 Chemistry Chapter 7 Equilibrium Short Question And Answers

Question 1. The unit of the equilibrium constant of the reaction, A + 3B ⇌ nC is L².mol^-2. What is the value of
Answer:

For the given \(K_c=\frac{[C]^n}{[A][B]^3}\)

Thus, the unit of

K_c = \(\frac{\left(\mathrm{mol} \cdot \mathrm{L}^{-1}\right)^n}{\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right) \times\left(\mathrm{mol} \cdot \mathrm{L}^{-1}\right)^3}\)

= (mol .L^-14)n^-4

Hence, L₂.mol^-2 = (mol . L^-1)n^-4

Or, n = 2

Read and Learn More NCERT Class 11 Chemistry Short Answer Questions

Question 2. Find out the value of K_p/K_c for the reaction \(\mathrm{PCl}_5(g) \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g)\) at 298K, consider the unit of concentration is mol L-1 and the unit of pressure is atm.
Answer:

We know. K_p = K_c(RT)An For the given reaction, Δn = (1 +1-1) = 1

Thus, \(K_p=K_c \times R T \quad \text { or, } \frac{K_p}{K_c}=R T\)

⇒ \(0.0821 \mathrm{~L} \cdot {atm} \cdot \mathrm{mol}^{-1} \mathrm{~K}^{-1} \times 298 \mathrm{~K}\)

= \(24.465 \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1}\)

NCERT Class 11 Chemistry Chapter 7 Short Questions and Answers

Question 3. How will the following reaction equilibrium be affected if the volume of each reaction system is increased at a constant temperature?
Answer:

At constant temperature, if the volume of the reaction system is increased, then the total pressure at equilibrium will decrease. Thus, the equilibrium of the system will be disturbed. According to Le Chatelier’s principle, equilibrium will be shifted in a direction that increases the total number of molecules.

In the first reaction, the equilibrium will shift to the left side. As a result, the product yield, [SO₃], will decrease. On the other hand, in the case of the second reaction, the equilibrium will shift to the right side, thereby increasing the yield of the product [CO(g)].

Question 4. Any reversible reaction’s equilibrium may be shifted to the left or reality changing the conditions. Will this change cause any alteration in the value of the equilibrium constant?
Answer:

At a certain temperature the equilibrium constant of a reversible reaction has a definite value.

If temperature remains fixed, then the equilibrium can be shifted to the left or right by changing the conditions of pressure, temperature, etc: on which the equilibrium of a reversible reaction depends. Consequently, the respective amount of both reactants and products will change, but the value of the equilibrium constant remains unchanged since the temperature remains fixed.

If the equilibrium is shifted due to temperature change, then the amounts of both reactants and products as well as the value of the equilibrium constant will change. With the increase in temperature, the value of the equilibrium constant will increase for an endothermic reaction and decrease for an exothermic reaction.

Question 5. At constant temperature, if the pressure is changed at the equilibrium of a gaseous reaction, then will the values of K_p, K_c, and K_x change?
Answer:

We known \(\Delta G^0=-R T \ln K_p \quad \text { or, } K_p=e^{-\frac{\Delta G^0}{R T}}\) where G° = standard free energy change ofthe reaction.

The value of ΔG° depends only on temperature. Its value is independent of pressure. So, the value of Kp is independent of pressures.

We know, K_p = K_c(RT)^Δn. Since the value of Kp does not depend on pressure, according to this relationship, the value of Kc is also independent of pressure.

Again, we know, K_p= K_x(p)n or \(K_x=\frac{K_p}{(P)^{\Delta n}}\)

As K_p does not depend on pressure, according to this relation, the value of Kx depends on pressure. However if Δn = 0, then pressure will not affect K_x.

Question 6. How can the yield of the products be increased by changing the volume of the reaction system in the given reactions at constant temperature?

C(s) + H₂O(g)⇌ CO(g) + H₂(g)

2H₂(g) + O₂ (g)⇌ 2H₂O(l)

Answer:

In this case, the volume increases in the reaction as written (since Δn = +1 ). Thus, if the volume of the reaction system is increased at a constant temperature, then the equilibrium will shift to the right, and consequently, the yields of the products will increase.

In this case, the volume decreases in the reaction as written (since Δn = -3 ). Thus, if the volume of the reaction system is decreased at constant temperature, then the equilibrium will shift to the right. As a result, the yield of products will increase.

Question 7. Mention two factors for which die yields of the products in the given reaction increase.

⇒ \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})  \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \text {-heat }\)

Answer:

Keeping both temperature and volume fixed, if we add some reactants [CO(g) or H₂O(g)] to the reaction system or remove some products [CO₂(g) or H₂(g) ] from the reaction system, equilibrium will shift to the right, which will result in higher yields of the products.

The reaction is endothermic. Thus, on increasing the temperature at equilibrium, the equilibrium of the reaction will shift to the right. This will cause higher yields of the products.

Question 8. What will be the change in concentrations of H₃O⁺ & OH^– and the ionic product of water (K_w) if NaOH is added to pure water at a certain temperature?
Answer:

Since Kw is fixed at a certain temperature, it will not undergo any change due to the addition of NaOH in pure water.

However, the concentration of OH- ions increases due to the addition of NaOH, causing the dissociation equilibrium of H₂O to shift to the left

⇒ \(\left[\mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)\right]\)

As a result, the concentration of H₃O+ ions in the solution is reduced

Question 9. 20 mL of 0.15(M) HCl solution is mixed with 50 mL of 0.1(M) CH₃COONa solution. State whether the mixed solution will act as a buffer or not.
Answer:

Number of millimoles of CH₃COONa₃ in 50 mL 0.1(M) CH₆COONa₃ =0.1 × 50 = 5 and that of HCl in 20mL 0.15(M)HCl= 0.15 × 20 = 3.

The reaction between CH₆COONa₂ and HCl is:

CH₃COONa(O + HCl(aq)→CH₃COOH(aq) + NaCl(aq)

Hence, 3 millimol of HCl + 3 millimol of CH₆COONa → 3 millimol of CH₃COOH + 3 millimol of NaCl.

Therefore, at the end of the reaction, there remains 3 millimol of CH₃COOH and (5-3) = 2 millimol of CH₃COONa.

∴ The resulting solution consists of weak acid (CH₃COOH) and its salt (CH₃COONa). So, it acts as a buffer.

NCERT Solutions Class 11 Chemistry Chapter 7 Short Q&A

Question 10. What would the effect on the yield of products be If the temperature of the following reaction systems Is changed at equilibrium?

N₂(g) + O₂(g) ⇌ 2NO(g); ΔH > 0

2SO₂(g) + O₂ (g) ⇌ 2SO₃(g); ΔH < 0

Answer:

The reaction is endothermic (as ΔH > 0 ). If the temperature is increased at the equilibrium of this reaction, then the equilibrium will shift to the right, and the yield of products will increase. On the other hand, if the temperature is reduced at the equilibrium of the reaction, then the equilibrium will shift to the left, and the yield of products will be reduced.

The reaction is exothermic (as ΔH < 0 ). If the temperature is decreased at the equilibrium of this reaction, then the equilibrium will shift to the right, and the yield of products will increase. On die other hand, if the temperature is increased at the equilibrium of the reaction, then the equilibrium will shift to the left, and the yield of products will be reduced.

Question 11. Identify Lewis acids and Lewis bases in the following reactions and give reasons:

1. SiF₄ + 2F→  SiF₆^2-

2. \(R M g X+2\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \ddot{\mathrm{O}}: {RMg}\left[\mathrm{O}\left(\mathrm{C}_2 \mathrm{H}_5\right)_2\right]_2 \mathrm{X}\)

3. \(\mathrm{Ag}^{+}+2 \ddot{\mathrm{N}} \mathrm{H}_3 \rightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}\)

4. \(\ddot{\mathrm{N}} \mathrm{H}_3+\mathrm{H}^{+} \rightarrow \stackrel{+}{\mathrm{N}} \mathrm{H}_4\)

Answer:

According to Lewis’s concept, an acid is a substance that can accept one or more electron pair(s). Generally, cations (such as Ag⁺, H⁺, K⁺), compounds having a central atom with an incomplete octet (such as SiF₄, AlF₃, RMgX, BF₃), and compounds whose central atom is linked to an electronegative atom by a double bond (such as GO₂ ) can act as Lewis acids. In the given reactions, Lewis acids are SiF₄, RMgX, Ag⁺, and H⁺.

According to Lewis’s concept, a base is a substance that can donate one or more electron pair(s). Anions (such as F^–, OH₂Si₂) and compounds with lone pairs of electrons can act as Lewis bases. Therefore, in the given reactions, Lewis bases are F-, NH₃, (C₂H₅)₂O, and NH₃.

Question 12. What will happen when a solution of potassium chloride is added to a saturated solution of lead chloride? Give reason.
Answer:

When potassium chloride solution is added to a saturated lead chloride solution, the solubility of lead chloride decreases due to the common ion (Cl^–) effect.

Explanation:

The following equilibrium is established in an aqueous PbCl₂ solution:

PbCl₂(s) Pb²⁺(ag) + 2Cl^– (ag)

The addition of KCl to the saturated solution of PbCl₂ increases the concentration of the common ion Cl^– the above equilibrium to get disturbed. To re-establish the equilibrium, some of the Cl^– ions will combine with an equivalent amount of Pb²⁺ ions to form solid PbCl₂. Therefore, as an overall effect, the equilibrium is shifted to the left. Hence, the solubility of PbCl₂ decreases.

Question 13. Why does not MgS04 form any precipitate when it reacts with NH₃ in the presence of NH₄Cl?
Answer:

NH₃ is a weak base. In an aqueous solution, it ionizes partially to produce NH+ and OH- ions.

⇒ \(\mathrm{NH}_3(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q)\)

In the presence of NH₄Cl, owing to the common ion effect of NH₄, the degree of ionization of NH₃ is further suppressed. Thus, the concentration of OH^– ions decreases to a large extent.

At this low concentration of OH^– ions, the product of the concentration of Mg²⁺ ions and square of the concentration of OH^– ions (as K_sp[Mg(OH)₂] =[Mg²⁺] × [OH^–]² cannot exceed the solubility product of Mg(OH)₂, i.e., [Mg²⁺] x [OH^–]² < K_sp (solubility product). As a result, Mg(OH)₂ does not precipitate.

Question 14. Will the pH of pure water at 20°C be lower or higher than that at 50°C?
Answer:

The ionic product of water (Kw) increases with, a temperature rise Hence

⇒ \(K_w\left(50^{\circ} \mathrm{C}\right)>K_w\left(20^{\circ} \mathrm{C}\right)\) or, \(p K_w\left(50^{\circ} \mathrm{C}\right)<p K_w\left(20^{\circ} \mathrm{C}\right).\)

Since pK_w=-log₂₀K_w

Now, for pure water \(p H=\frac{1}{2} p K_w\)

Question 15. Both CuS and ZnS are precipitated if H₂S gas is passed through an alkaline solution of Cu²⁺ and Zn²⁺. Explain.
Answer:

In an aqueous solution, H₂S ionizes to establish the following equilibrium,

H₂S(aq) + 2HzO(l) ⇌ 2H₃O⁺(aq) + S^2-(aq)

The degree of ionization of H₂S increases in alkaline solution because OH^– ions present in the solution react with H₃O⁺ ignite) form unionized water molecules. This shifts the equilibrium to the right, thereby increasing the concentration of S^2- ions. in the presence of a high concentration of S^2- ions, [Cu²⁺] (S^2-] > and [Zn²⁺][S^2-] > Ksp(ZnS). As a result, both CuS and ZnS are precipitous.

Question 16. Why is an aqueous solution of NaNO₃ neutral?
Answer:

NaNO₃ is a salt of strong acid HNO₃ and strong base NaOH. In its aqueous solution, NaNO₃ dissociates completely, forming Na+ and NO₃ ions. In an aqueous solution, Na⁺(aq) is a weaker acid than H₂O and NO₃ is a weaker base than H₂O.

So, neither Na+(a₂) nor NO₂(aq) reacts with water. As a result, there is no change in the concentration of either H₃O⁺ ions or OH^– ions. Due to this, an aqueous solution of NaNO₃ is neutral.

Question 17. An aqueous solution o/(NH4)₂SO₄ is acidic. Explain
Answer:

(NH₄)₂SO₄ is a salt of a weak base (NH₃) and a strong acid (H₂SO₄). In its aqueous solution, (NH₄)₂SO₄ dissociates almost completely forming NH₄⁺ and SO₄^2- ions. SO₄^2- ion is a conjugate base of strong acid H₂SO₄ and hence in aqueous solution, it is a very weak base in comparison to H₂O.

As a result, the SO₄^2- ion does not react with water in aqueous solution. On the other hand, NH₄ is a conjugate acid of weak base NH₃. In an aqueous solution, the NH₄⁺ ion shows a higher acidic character than H₂O.

As a result, NH₄ ions react with water.

[NH₄(O^-7) + H₂O(l)⇌NH₃(aq) + H₃O+(aq)]

Causing an increase in the concentration of H₃O⁺(aq) ions in the solution. This makes the solution of (NH₄)₂SO₄ acidic.

Question 18. At 25°C what is the concentration of H₃O⁺ ions in an aqueous solution in which the concentration of OH^– ions is 2 × 10^-5(M)?
Answer:

At 25 °C, K_w = 10-14. Now, for an aqueous solution, [H₃O+] × [OH^–]

= K_w At 25 °C, [H₃O⁺][OH^–] = 10^-14

Given: [OH^–] = 2 × 10^-5(M)

Therefore, [H₃O⁺] \(=\frac{10^{-14}}{2 \times 10^{-5}}=5 \times 10^{-10}(\mathrm{M})\)

Question 19. At a certain temperature, the ratio of ionization constants of weak acids HA and HB is 100:1. The molarity of the solution is the same as that of HB, and the degrees of ionization of HA and HB in their respective solutions are α₁ and α₂ respectively, then show that α₁ = 10α₂
Answer:

Suppose, the ionization constants of HA and MB arc K₁ and K₂, respectively. If the concentration of each of the solutions of HA and HB is c mol.L^-1, then

⇒ \(\alpha_1=\sqrt{\frac{K_1}{c}} \text { and } \alpha_2=\sqrt{\frac{K_2}{c}} \text {; }\)

Where α₁ and α₂ are the degrees of ionization of HA and HB in their respective solutions.

It is given that K₁ : K₂ = 100: 1

⇒ \(\text { So, } \frac{\alpha_1}{\alpha_2}=\sqrt{\frac{K_1}{K_2}}=\sqrt{100}=10 \text {, i.e., } \alpha_1=10 \alpha_2\)

Question 20. Find [OH^–] in pure water if [H₃O⁺] in it is x mol L^-1. Also, find the relation between x K_w.
Answer:

In pure water,

⇒  \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]\)

Therefore, [OH^– ] in pure water \(=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=x \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

We know, Kw = [H₃O⁺] × [OH^–]

∴ K_w = x × x

or, K_w = x²

∴ \(x=\sqrt{K_w}\)

Class 11 Chemistry Chapter 7 Equilibrium Short Answers

Question 21. Identify the Lewis acids and Lewis bases in the following reactions

1. H⁺ + OH^–→ H₂O
2. Co³⁺ + 6NH₃ ⇌  [CO(NH₃)6)³⁺
3. BF₃ + :NH₃→ [H₃N→ BF₃]
4. CO₂ + OH^–→HCO₃^–
5. AlF₃ + 3F^–→AlFl₆^3-

Answer:

According to Lewis’s concept, an acid is a substance that can accept a pair of electrons. Generally, cations 

Example: Ag⁺, H⁺, K⁺, etc.

Molecules with the central atom having incomplete octet

Example: SiF₄, AlF₃, RMgX, BF₂,

And molecules in which the central atom is linked to an electronegative atom through double bonds

Example: CO₂  can act as Lewis acid. In the given reactions, Lewis acids are H⁺, CO³⁺, BF₃, CO₂, and AlF₃.

According to Lewis’s concept, a base is a substance that can donate a pair of electrons. Anions

Example; F^–, OH^–, etc.) and molecules having unshared electron pairs act as Lewis bases. In the given reactions, Lewis bases are OH^–,: NH₃, F^–

Question 22. We know, ΔG° = -RTInKc and ΔG° = -RTlnKp. Therefore in case of a reaction occurring in the gaseous phase at a given temperature, ΔG° is the same even if the values of K_p and K_c are different. Is the statement true? Give reasons.
Answer:

The statement is not true. In the equation, ΔG° = -RTnKc, the concentration of each of the reactants and products at standard state is taken as l(M).

On the other hand, in the equation, ΔG° = -RTnKp, the partial pressure of each of the reactants and products at standard state is taken as 1 atm. Therefore, the different values of AG° will be obtained from these two equations.

The values of the equilibrium constant (K) of a reaction at 25°C and 50°C are 2 × 10^-1 and 2 ×10^-2 respectively. Is the reaction an exothermic or endothermic?

Question 23. Consider the reaction, \(2 \times a=c \text { or, } a=\frac{c}{2} \text {. }\) Heat and answer the following questions:

1. Find the relation among a, b, and c.
2. State whether the equilibrium will be shifted towards right or left if the temperature is Increased.

Answer:

According to the equation ofthe reaction, c mol of XY forms when a mol of X₂ reacts with b mol of Y₂. So, the number of X atoms in a mol of X₂ = The number of X atoms in c mol of XY.

since \(2 \times a=c \text { or, } a=\frac{c}{2} \text {. }\)

Similarly, the number of Y atoms in b mol of Y₂ = the number of Y atoms in c mol of XY.

∴ \(2 \times b=c \text { or, } b=\frac{c}{2}\)

∴ \(a=b=\frac{c}{2} \text {. }\)

The reaction is exothermic. So, according to Le Chatelier’s principle, a temperature rise will cause the equilibrium ofthe reaction to shift to the left.

Question 24. State Le Chatelier’s principle, explain the effect of (a) pressure and (b) continuous removal at the constant temperature on the position of equilibrium of the following reaction:

H₂(g) + I₂(g) ⇌ 2HI(g)

Answer:

Pressure does not have any effect on the position of the equilibrium because the reaction is not associated with any volume change [Total no. of molecules of HI(g) = Total no. of molecules of H₂(g) and I₂(g)].

If HI is removed continuously from the reaction system, then the equilibrium goes on shifting towards the right, and finally, the reaction moves towards completion.

Question 25. Consider the following reaction:

2A(g) + B₂(g) ⇌ 2AB(g); ΔH < 0. How can the yield of AB(g) be Increased?
Answer:

For the reaction ΔH < 0, it is an exothermic reaction. According to Le Chatelier’s principle, if the temperature of an exothermic reaction at equilibrium is decreased, the equilibrium of the reaction shifts to the right. So, the decrease in temperature will result in a higher yield of AB(g).

The given reversible reaction is associated with the decrease in number of moles [An = 2- (2 + 1) =-l ] in the forward direction. So, according to Le Chatelier’s principle, if the pressure is increased at the equilibrium of the reaction, the equilibrium shifts to the right, thereby increasing the yield of AB(g)

At constant temperature and volume, if the reactant A₂(g) or B₂(g) is added to the reaction system at equilibrium, then, according to Le Chatelier’s principle, the equilibrium of the reaction will shift to the right. As a result, the yield of AB(g) will increase.

Short Questions for Class 11 Chemistry Chapter 7 Equilibrium

Question 26. At a particular temperature, the following reaction is carried out with 1 mol of A(g) and 1 mol of B(g) in a closed vessel:

A(g) + 4B(g) ⇌ AB₄(g). Will the equilibrium concentration of AB₄(g) be higher than that of A(g)?
Answer:

According to the given equation of the reaction, 1 mol of AB₄ forms due to the reaction between 1 mol of A and 4 mol of B. Suppose, the concentration of AB₄ at the equilibrium of the reaction is x mol- L^-1

So, according to the given equation, the concentration of A and B at equilibrium will be (1 – x) mol- L^-1 and (1-4) mol. L^-1 respectively. At equilibrium, if the concentration of AB, was greater than that of A, then x would be greater than (1 – x), i.e., x > 1 – x or, x > 0.5.

If x was greater than 0.5, then the concentration of B would be negative. This is impossible. Therefore, the concentration of AB₄ can never be greater than that of A.

Question 27. For the reaction 2H₂(g) + O₂(g) ⇌  2H₂O(g) — K_p = K_c(RT)^x. Find the value of.
Answer:

For the given reaction, Δn = 2-(2+1) = -1

Δn = 2-3 = -1

We know, K_p = K_c(RT)^Δn

As Δn = -1, K_p = K_c(RT)^-1 …………………(1)

Given, K_p = K_c(RT)^x…………………(2)

Comparing equations (1) and (2), we have x = -1

Question 28. At 200°C, the equilibrium N₂O₄(g) 2NO₂(g) is achieved through the following two pathways: 0.1 mol N₂O₄ is heated in a closed vessel L volume, A mixture of 0.05 mol N₂O₄(g) and 0.05 mol NO₂(g) is heated at 200°C in a closed vessel of 1 L volume. In these two cases, will the equilibrium concentrations of N₂O₄(g) and NO₂(g) and the values of equilibrium constants be the same?
Answer:

The value of the equilibrium constant of a reaction depends only on temperature. It does not depend on the initial concentrations of the reactants. Since the temperature is the same for both experiments, the value of the equilibrium constant will be the same in both cases.

The initial concentrations of the reactant(s) in the two experiments are not the same. As a result, the molar concentrations of N₂O₄(g) and NO₂(g) at equilibrium will be different in the two experiments.

NCERT Class 11 Chemistry Chapter 7 Equilibrium Solutions

Question 29. What will be the relation between Kp and Kc for the given equilibrium?CO(g) + H₂O(g )⇒ CO₂(g) + H₂(g)
Answer:

We know, \(K_p=K_c(R T)^{\Delta n}\)

For the given reaction, An = (1 + 1) – (1 +1) = 0

Therefore, for the given reaction K_p = K_c(RT)° – K_c

Question 30. At a given temperature, for reaction A B, the rate constant (k) of the forward reaction is greater than that of the backward reaction (kb). Is the value of the equilibrium constant (K) for this reaction greater than, less than, or equal to 1?
Answer:

The equilibrium constant (K) of a reaction

⇒ \(=\frac{\text { Rate constant of the forward reaction }}{\text { Rate constant of the backward reaction }}=\frac{K_f}{K_b}\)

Given: Kj→ Kb. Therefore K > 1.

CBSE Class 11 Chemistry Notes For Chapter 7 Equilibrium Introduction

Under a given set of experimental conditions, a system is said to be at equilibrium if the macroscopic properties of the system, such as temperature, pressure, concentration, etc. do not show any change with time.

There are two types of equilibria:

Static equilibrium and Dynamic equilibrium. Equilibrium involving physical and chemical changes is dynamic.

A dynamic equilibrium is established when two or more opposing processes occur in a system at the same rate. For example, if the decomposition of hydrogen iodide [2HI(g)⇌ H₂(g) +I₂(g)] is carried out in a closed vessel,  it is found that the reaction is never complete.

At the onset of the reaction, the system contains only hydrogen iodide (reactant) molecules. With time, the concentration of molecules gradually decreases.

In contrast, the concentrations of H₂ and I₂ (product) molecules gradually increase till a stage is reached at which no further change in concentrations of either the reactants or the products takes place.

At this stage, the reaction appears to have stopped. This state of the system at which no further change occurs is called a state of equilibrium. This state of equilibrium is not static, but it is dynamic because the forward and backward reactions are still going on at the same rate.

Due to this dynamic nature of equilibrium, no change in concentration and other properties of the system occurs at the equilibrium state.

CBSE Class 11 Chemistry Notes Chapter 7 Equilibrium

Read and Learn More CBSE Class 11 Chemistry Notes

Equilibrium involving chemical reaction (i.e., chemical equilibrium) is represented as:

Reactants; F=± Products The double half arrows indicate that the reactions in both directions are going on simultaneously.

The mixture consisting of reactants and products in the equilibrium state is called an equilibrium mixture. Dynamic equilibrium is also observed in case of physical changes, particularly during the transition of state example melting of solids, vaporization of liquids, etc.

CBSE Class 11 Chemistry Notes For  Physical Equilibrium

Equilibrium involving physical processes is called physical equilibrium. Thus, the equilibria attained during the dissolution of a salt, the evaporation of a liquid, etc., are examples of physical equilibria. Different types of physical equilibria are briefly discussed in the following section.

Solid-liquid Equilibrium

At the melting point (or freezing point) of a pure substance, both its solid and liquid phases co-exist and a dynamic equilibrium develops between the two phases:

Solid-liquid When the system with the above equilibrium mixture is heated, the temperature of the system remains constant until the whole solid transforms into liquid. Similarly, if heat is withdrawn from this system, the temperature of the system remains constant until the whole liquid transforms into a solid.

Melting point Or freezing point

At Normal atmospheric pressure, the temperature at which the solid and the liquid states of a pure substance remain in equilibrium is called the normal melting point (or normal freezing point) of the substance.

When the solid and liquid phases of a pure substance are kept in contact with each other at its melting point in a closed insulated container, no exchange of heat takes place between the system and its surroundings.

However, a state of dynamic equilibrium is established between the solid and the liquid phases inside the container. It is also observed that the masses of solid and liquid phases do not change with time and the temperature of the system remains constant.

Example: Let us take some ice cubes together with some water inside a thermos flask at 0°C and 1 atm pressure and leave the mixture undisturbed. After some time it will be seen that the masses of ice and water are not changing with time and also the temperature remains unchanged. This represents an equilibrium between ice and water

Equilibrium: H₂O (s)⇌H₂O(l)

• Although we observed apparent change inside the thermos flask, a careful examination shows that some activity is still going on between the two phases of water.
• Some molecules of ice convert into water, while at the same time, the same number of molecules of water convert into ice.
• However, as the masses of ice and water remain unchanged, it can be concluded that the two opposite processes (i.e., melting of ice and freezing of water) occur at the same rate.

The rate of melting of ice = The rate of freezing of water

Thus, the equilibrium that is established in the solid-liquid system is dynamic.

Liquid-vapour equilibrium

• The equilibrium between a liquid and its vapor can be better understood if we consider the vaporization of water in a closed vessel. Let us take a closed vessel connected to a manometer and a vacuum pump as shown in
• The closed vessel is first evacuated. The levels of mercury are the same in both the limbs of the manometer.
• Then some pure water is introduced into the vessel and the whole apparatus is kept at room temperature (or any desired temperature by placing it in a thermostat).
• After some time it is seen that the level of mercury in the left limb begins to fall and the right begins to rise and eventually the levels of mercury in both limbs become fixed at two different levels. Under this condition, the system is said to have attained equilibrium

Equilibrium: H₂O(Z) ⇌H₂O(g)

Class 11 Chemistry Chapter 7 Equilibrium NCERT Notes

Molecular interpretation:

At the initial stage of the experiment, as more and more water changes into vapor (by evaporation), the pressure inside the vessel gradually increases. This is indicated by the fall in mercury level in the left limb of the manometer.

• The molecules of water vapor, so produced, collide among themselves, with the walls of the vessel and also with the surface of the water.
• Water vapor molecules with lower kinetic energy get converted into liquid states when they come in contact with the surface of water. This is called condensation.
• At the beginning of the experiment, the rate of evaporation of water is greater than the rate of condensation of its vapor. However, with time, the rate of condensation increases, and that of evaporation decreases.
• After some time, the rates of evaporation and condensation become equal. It is said that a dynamic equilibrium is established between water and its vapor

At equilibrium: The rate of evaporation = The rate of condensation

• The difference in the levels of mercury in the two limbs gives a measure of the equilibrium vapor pressure or saturated vapor pressure of water at the experimental temperature.
• At a fixed temperature, if a liquid remains in equilibrium with its vapor, then the pressure exerted by the vapor is called the vapor pressure of the liquid at that temperature.
• An equilibrium between a liquid and its vapor is established only in a closed vessel. If the liquid is placed in an open vessel, then its vapor diffuses into the air. Consequently, no equilibrium is established between a liquid and its vapor in an open container.

Solid Vapour equilibrium

In general, solid substances have verylow vapor pressure compared with liquid substances at the same temperature.

• However, some substances, e.g., iodine, camphor, solid CO₂, naphthalene, etc. have high vapor pressure even at ordinary temperatures.
• Such substances can convert directly from the solid to the vapor state without passing through the liquid state.
• In this process of transformation of a solid directly to the When sublimation of volatile solids done in a closed vessel, equilibrium is established between the solid and its vapor.
• Example: If we take some solid iodine [I₂(s)] in a closed vessel and heat it below its melting point (113.6°C), it is found that the vessel gets filled with violet vapor of iodine.
• Initially the intensity of color increases and eventually it becomes constant. Under this condition, the rate of sublimation of solid iodine is equal to the rate of condensation of iodine vapor. This results in a state of dynamic equilibrium as below,

Equilibrium: I₂(s)⇌ I₂(g)

Other substances showing this kind of equilibrium are:

• NH₄Cl(s)⇌NH₄Cl(g)
• Camphor (s)⇌ Camphor (g)

Equilibrium involving dissolution of solid in liquid

• Suppose, at a fixed temperature, an excess amount of a solid substance, say sugar (solute), is added to a definite volume of a suitable solvent (say water) taken in a beaker and then the mixture is stirred well with a glass rod.
• The particles (i.e., molecules in case of non-electrolytes and ions in case of electrolytes) of solute gradually pass into the solvent, thereby increasing the concentration of the solute in the solution. This process is called dissolution of solute.
• Then a stage comes when no more solute dissolves in the solvent. Instead, the solute settles down at the bottom of the beaker i.e., a saturated solution is obtained.
• During the process of dissolution of solute, the reverse process also occurs simultaneously, i.e., the solute particles from the solution get deposited on the surface of the undissolved solute (a process called crystallization).
• Initially, the rate of dissolution of the solid solute is higher than the rate of crystallization of the dissolved solute.
• However, with time, as the solution becomes more and more concentrated, the rate of dissolution decreases, and that of crystallization increases. Finally, the rate of dissolution becomes equal to that of crystallization.
• Under this condition, a state of equilibrium is established between the dissolved solute particles and the undissolved solid solute. The equilibrium can be represented as,

Solute (solid)⇌Solute (in solution)

This state of equilibrium is said to be dynamic because the process of dissolution and crystallization continues as long as the temperature and other external conditions remain unchanged.

The solution obtained at equilibrium is called a saturated solution. The concentration of the saturated solution depends on the temperature.

Equilibrium involving the dissolution of gas in liquid

1. The solubility of a gas in a given liquid depends on the experimental temperature and pressure and also on the nature of the liquid and the gas under consideration.
2. When a gas (say CO₂) comes in contact with a liquid, the molecules of the gas begin to collide with the surface of the liquid. Consequently, some of the gas molecules get attracted by the molecules of the liquid and ultimately pass into the liquid phase.
3. % At a fixed temperature and pressure, if a gas is passed continuously through a fixed amount of a liquid kept in a closed vessel, gas molecules get dissolved in the liquid, and eventually, a saturated solution of the gas in the liquid is obtained.
4. In this solution, a dynamic equilibrium is established between the dissolved gas and the gas over the liquid surface.
5. Under this condition, the rate of dissolution of the gas molecules in the liquid is equal to the rate at which the dissolved gas molecules escape from the solution.
6. Thus, it is a dynamic equilibrium; Liquid + Gas Dissolved gas.

Taking CO₂ as the gaseous substance it can be represented as:

At a fixed temperature, the amount of gas dissolved in a liquid depends upon the pressure of the gas over the liquid. The concentration of the dissolved gas increases with the increase in the pressure of the gas. Henry’s law demonstrates how the solubility of a gas in a liquid varies with pressure at a given temperature.

Henry’s law:

The mass (or mole fraction) of a gas dissolved in a given mass of a solvent at a given temperature is directly proportional to the pressure of the gas over the solvent.

In a fixed amount of a liquid, if a gas with a pressure of p dissolves by an amount of w, then according to Henry’s law, wp or, w = kp [k is the proportionality constant]

The reason for fizzing out of CO₂ gas when a soda water uc is exposed:

In a sealed soda water bottle, CO₂ remains dissolved in liquid under high pressure, and an equilibrium exists between the dissolved CO₂ and CO₂ gas present over the liquid.

• As soon as the bottle is opened, the pressure of CO₂ gas over the liquid decreases and becomes equal to the atmospheric pressure.
• Since the solubility of a gas in a liquid is proportional to the pressure of the gas, the solubility decreases considerably because of the lowering of pressure.
• As a result, a large amount of dissolved CO₂ escapes from the solution until a new equilibrium is established.

This phenomenon of escaping CO₂ gas is associated with a fizzing sound. This is also the reason why a soda water bottle turns flat when left open in the air for some time.

CBSE Class 11 Chemistry Notes For Irrversible And Reversible Reactions

Irreversible reactions

A reaction in which the products formed do not react together to revert to the reactants despite the changes in reaction conditions is called an irreversible reaction.

Examples: When potassium chlorate (KClO₃) is heated in an open vessel, potassium chloride (KCl) and oxygen (O₂) are produced.

However, KCl and O₂ do not react with each other to regenerate KClO₃. So, the thermal decomposition of KClO₂ is an example of an irreversible reaction.

Most of the ionic reactions are irreversible. For example, when an aqueous solution of KCl is treated with an aqueous AgNO₃ solution, a curdy white precipitate of AgCl and KNO₃ is produced, but the precipitated AgCl and KNO₃ do not react back to AgNO₃ and KCl.

⇒ \(\mathrm{AgNO}_3(a q)+\mathrm{KCl}(a q) \rightarrow \mathrm{AgCl}(s) \downarrow+\mathrm{KNO}_3(a q)\)

Characteristics of an irreversible rea

ction:

1. The products in an irreversible reaction do not show any tendency to react together. So, the reaction in the opposite direction can never happen. For this reason, an irreversible reaction attains completion in course of time.
2. Since an irreversible reaction undergoes completion, the reactants participating in the reaction in equivalent amounts are completely exhausted.
3. Irreversible reactions are accompanied by a decrease in Gibbs free energy (i.e., ΔG<0).

Reversible reactions

A Reaction in which the products formed react together to regenerate the reactants, and an equilibrium is established between the reactants and products under the condition of the reaction is called a reversible reaction.

Example: The thermal decomposition of NH₄Cl vapor in a closed vessel is a reversible reaction.

⇒ \(\mathrm{NH}_4 \mathrm{Cl}(\text { vapour }) \rightleftharpoons \mathrm{NH}_3(g)+\mathrm{HCl}(g)\)

Explanation:

When NH₄Cl(vapor) is heated in a closed vessel at 350°C, it undergoes thermal decomposition, producing NH₃ and HCl gases. However, even after a long time, it is observed that the reaction mixture contains not only NH₃ and HCl but also NH₄Cl vapor.

This proves that the decomposition of NH₄Cl vapor in a closed container never gets completed. In another closed vessel, if an equimolar mixture of NH₃ and HCl gases is heated at 350°C for a long time, the vessel is found to contain NH₄Cl vapor along with NH₃ and HCl gases.

This means that the reaction between NH₃ and HCl in a closed vessel never gets completed. Thus, it can be concluded that on heating NH₄Cl vapor, it decomposes to produce NH₃ and HCl gases which again react partially to form NH₄Cl vapor.

Hence, the thermal decomposition of NH₄Cl vapor is a reversible reaction. The following reactions show reversibility when carried out in a closed vessel.

⇒ \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) ; \mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)\)

⇒ \(\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g) ; \mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)\)

Characteristics of reversible reaction:

In a reversible reaction, both the forward and the backward reactions occur simultaneously. In the forward reaction, the reactants react together to yield the products, while the products react together to produce the reactants in the backward reaction.

NCERT Solutions Class 11 Chemistry Chapter 7 Equilibrium

For example, when an equimolecular mixture of H₂ gas and I₂ vapor is heated in a closed container, the following reaction takes place—

⇒ \(\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) .\)

Here, the forward reaction is: H₂(g) + I₂(g) →2HI(g) and the backward reaction is: 2HI(g)→H₂(g) + I₂(g)

Since a reversible reaction is not complete, the reactants are not completely consumed in such reactions. Instead, a mixture containing both the reactants and the products is obtained.

• Such reactions achieve an equilibrium state when the rate of the forward reaction becomes equal to that of the backwaed reaction.
• At a given temperature and pressure, when a reversible reaction reaches equilibrum, the gibbs free energy change becomes zero i.e., ΔGp, T=0
• Reversibility and irreversibility of chemical reactions when carried out in open and closed containers
• Many chemical reactions that are found to be irreversible when carried out in open containers become reversible when they are carried outin closed containers.
• Different results are obtained when solid calcium carbonate (CaCO₃) is heated separately in a closed container and in an open container.

Explanation: On strong heating solid CaCO₂ decomposes to solid CaO and CO₂ gas. If CaCO₃ is decomposed in an open container, CO₂ gas escapes from the container into the air, and only solid CaO remains as residue. As the reactant (solid CaCO₃) in this reaction gets converted into products completely, the reaction is considered as an irreversible reaction.

H₂(g) +I₂(g) ⇌ 2HI(g).

If the same quantity of CaCO₃ is decomposed in a closed container, some quantity of CaCO₃ is still found to remain undecomposed. This is because CO₂ produced in the reaction cannot escape from the container.

As a result, a portion of CO₂ gas reacts with an equivalent amount of CaO to form CaCO₃ again. Hence, in the closed vessel, the reaction occurs reversibly. Consequently, a mixture of CaCO₃, CaO, and CO₂ is found to be present in the reaction vessel.

⇒ \(\mathrm{CaCO}_3(\mathrm{~s}) \stackrel{\Delta}{\rightleftharpoons} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

When steam is passed over the red-hot iron, ferrosoferric oxide (Fe₃O₄) and H₂ gas are produced.

Explanation: If the reaction is carried out in an open vessel, the H₂ gas produced diffuses into the air. So, Fe₃O₄ resulting from the reaction cannot have hydrogen gas to react with.

Consequently, the reverse reaction cannot take place. For this reason, at the end of the reaction, only Fe₃O₄ is left behind as residue in the reaction vessel.

⇒ \(3 \mathrm{Fe}(s)+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Fe}_3 \mathrm{O}_4(s)+4 \mathrm{H}_2(\mathrm{~g}) \uparrow\)

However, if the reaction is carried out in a closed vessel, the H₂ gas produced cannot escape from the container. As a result, a certain amounts of H₂ gas and Fe₃O₄ together and regenerate Fe and H₂O.

Hence, in a closed vessel, the reaction occurs both in the forward and reserve. directions, giving a mixture of Fe(s), H₂O(g), Fe₃O₄(s) and H₂(g).

⇒ \(3 \mathrm{Fe}(s)+4 \mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_3 \mathrm{O}_4(s)+4 \mathrm{H}_2(g)\)

CBSE Class 11 Chemistry Notes For  Chemical Equilibrium

Let us consider a hypothetical reversible reaction A +B⇌C+D, Which is started with 1 mol of A and lmol of B in a closed container at a given temperature.

1. At the beginning, the reaction system does not contain C and D (products). It contains only A and B (reactants). So, the reaction occurs only in the forward direction (A + B→C+ D).

2. At the outset of the reaction, since the concentrations of the reactants are maximum, the rate of the forward reaction is also maximum. This is because the rate of a reaction is directly proportional to the concentrations of the reactants.

3. As there are no C and D molecules at the start, the backward reaction (C+D→A + B) does not occur. However, the backward reaction starts occurring with the formation of A and B in the forward reaction. As C and D accumulate is the reaction system, they begin to react together to form A and B.

4. With time, the concentrations of C and D increase, while the concentrations of A and B decrease. As a result, the rate of the backward reaction increases, while that of the forward reaction decreases.

5. Eventually, a moment comes when the rate of the forward reaction becomes equal to the rate ofthe backward reaction.

6. When the rate of the forward reaction is equal to that of the reverse reaction, the reaction is said to have reached the state of equilibrium. However, at equilibrium, the reaction does not stop; instead, both the forward and backward reactions occur simultaneously at the same rate

7. The mixture of reactants and products at the equilibrium of a reaction is called the equilibrium mixture. The concentrations of the reactants and products in the equilibrium mixture are called their equilibrium concentrations. If the conditions {i.e., temperature, pressure, etc.) of the reaction remain undisturbed the relative concentrations of the reactants and products in the equilibrium mixture do not change with time.

Chemical equilibrium

The state of a reversible chemical reaction at a given temperature and pressure when the rates of the forward and reverse reactions become the same, and the concentrations of the reactants and products remain constant with time then the particular state is called the state of chemical equilibrium.

Chemical equilibrium is a dynamic

After the attainment of equilibrium of a reversible chemical reaction, if the reaction system is left undisturbed for an indefinite period at constant temperature and pressure,

• Then the relative amounts of the reactants and the products are found to remain unaltered.
• This observation leads to the impression that a reaction stops completely at equilibrium. However, it has been proved experimentally that the reaction does not cease rather both the forward and the backward reactions continue at the same rate. This is the reason why chemical equilibrium is designated as a dynamic equilibrium.
• Experimental proof of the dynamic nature of chemical equilibrium:
• When some quantity of pure CaCO₃ is heated strongly above 827°Cin a closed vessel (A), it decomposes into CaO and CO_2, and an equilibrium is established

⇒ \(\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

At equilibrium, the temperature and pressure of the reaction vessel remain unaltered with time.

• Now this vessel is connected with another closed vessel 14 containing CO₂ at the equilibrium pressure and temperature in such a way that there will be no effect on the equilibrium of the reaction occurring in vessel (A).
• After some time, a small quantity of solid is collected from the vessel (A) and analyzed. The analytical data indicates the presence of 14C in CaCO₃.
• This is possible only if some amount of 14CO₂ and CaO combine to form Ca14CO₃ at equilibrium. At the same time, some quantity of CaCO₃ decomposes to produce CaO and CO₂.
• So, the pressure on the reaction vessel remains constant. Thus, this experiment proves that even after the attainment of equilibrium, the reactions do not cease, both the forward and the reverse reactions proceed simultaneously at the same rate.

Characteristics of chemical equilibrium

Permanency of chemical equilibrium:

As long as the conditions under which a reaction attains equilibrium remain unaltered, no further change in equilibrium takes place, that is to say, the composition of the equilibrium mixture and other properties of it remains the same with time.

Dynamic nature of equilibrium:

Even after the attainment of equilibrium, a chemical reaction does not cease; both the forward and the reverse reactions continue at equal rates.

Incompleteness of the reaction at equilibrium:

At the equilibrium of a reaction, both the forward and reverse reactions take place simultaneously at the same rate. If any one of these reactions goes to completion, then the term equilibrium becomes irrelevant. Hence, for the equilibrium to exist, the reactions of both directions will have to be incomplete.

Approachability of equilibrium from either direction:

Under a given set of conditions, a reversible reaction attains the same equilibrium state irrespective of whether the reaction is started with its reactants or products.

Example: H₂ gas and I₂ vapor are allowed to react with each other in a closed vessel at 445°C. Eventually, the following equilibrium is established,

⇒ \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

The equilibrium mixture is found to contain H₁, H_2, and I₂ with mole percents of 80%, 10%, and 10%, respectively. In a separate container with the same volume, if 2 moles of H₁ gas are heated at 445°C, then H₁ gas decomposes to H₂ and I₂ gases, and eventually, the following equilibrium is established, 2HI(g) H₂(g) + I₂(g).

Here also, the equilibrium consists of H₁, H_2, and I₂ with mole percents of 80%, 10%, and 10%, respectively. Thus, the same equilibrium mixture is obtained, no matter whether we start the reaction with HI(g) or H₂(g) and l₂(g).

Chemical Equilibrium Class 11 Chemistry Notes

A catalyst cannot alter the state of equilibrium:

A catalyst is a substance that enhances the rate of a reaction without being used up in the reaction. A catalyst does not affect the position of equilibrium in a reaction. Its only function is to reduce the time that a reaction takes to reach an equilibrium state.

In the presence of a catalyst, the forward and reverse reactions of a reversible reaction are speeded up to the same extent. Under a given set of conditions, if a reaction is carried out in the presence or absence of a catalyst, then the same equilibrium mixture is obtained. That is, in both cases, the concentrations of the reactants and products are found to be the same.

Homogeneous and heterogeneous equilibria

Homogeneous equilibrium:

An equilibrium in which all the substances, Z.e., reactants, and products, are in the same phase is known as homogeneous equilibrium.

Examples 1. N₂(g) + 3H₂(g)⇌ 2NH₃(g)

2SO₂(g) + O₂(g) ⇌2SO₃(g)

CH₃COOH(Z) + C₂H₅OH(l) ⇌CH₃COOC₂H₅(Z) + H₂0(l)

Heterogeneous equilibrium:

An equilibrium in which the reactants and products are in different phases is known as heterogeneous equilibrium.

Examples CaCO₃(s) ⇌CaO(s) + CO₂(g)

2HgO(s)⇌ 2Hg(l) + O₂(g)

CBSE Class 11 Chemistry Notes For  The Law Of Mass Action

In 1864, C.W. Guldberg and P. Waage formulated a law regarding the dependence of the reaction rate on the concentration of the reactant. This law is known as the law of mass action.

At a constant temperature, the rate of a chemical reaction at any instant during the reaction is directly proportional to the active mass of each of the reactants at that instant.

So, the rate of a reaction increases with the increase in active masses of the reactants, while it decreases with the decrease in active masses of the reactants.

Active mass:

The active mass of a substance is generally considered as the same as its molar concentration. The active mass is expressed in different ways.

• In case of a dissolved substance is a solution, the active mass of the substance is taken to be the same as its molar concentration.
• If a VL solution contains n mol of a substance, then the active mass (or molar concentration) of the substance is n/v .
• In the case of a component gas in a gas mixture, the active mass of the component can be expressed either in terms of its molar concentration or partial pressure in the mixture.
• This is because the partial pressure of a component gas in a gas mixture is directly proportional to its molar concentration.

For a pure solid or liquid, the active mass is always taken as unity (1).

The molar concentration of a pure solid or liquid is directly proportional to its density:

⇒ \(\frac{\text { number of moles of the substance }}{\text { volume of the substance (in } \mathrm{L})}\)

= \(\frac{\text { mass of the substance }}{\text { molar mass of the substance }} \times \frac{1}{\text { volume of the substance (in L) }}\)

= \(\frac{\text { mass of the substance }}{\text { volume of the substance (in } \mathrm{L})} \times \frac{1}{\text { molar mass of the substance }}\)

= \(\frac{\text { density of the substance }}{\text { molar mass of the substance }}\)

As the molar mass of a pure substance is a fixed quantity, the molar concentration of a pure solid or liquid is directly proportional to its density. The density of a pure solid or liquid is constant at a given temperature, so its molar concentration.

Mathematical expression of the law of mass action: Let us consider the following simple chemical reaction in which one mole of A reacts with one mole of B, forming one:

Mole of C: A + B→C According to the law of mass action, at a particular moment during the reaction, the rate of the reaction, or, r = k(A) (B) Where (A) and (B) are the active masses or molar concentration + of A and B, respectively at that moment, and k is proportionality constant, known as the rate constant of the reaction. Equation (1) represents the rate equation of the said reaction. Here is a table is which some reactions and their rate equations are given.

General statement of the law of mass action:

At constant temperature, the rate of a chemical reaction at any instant is directly proportional to the product of molar concentrations (active masses) of the reactants at that instant, each concentration (active mass) term being raised to a power which appears as a stoichiometric coefficient of the species in the balanced chemical equation of the reaction.

Mathematical Form Of The Law Of Mass Action For A Reversible Reaction

Suppose, a reaction is started with ‘ a’ mol of A and ‘b 1 mol of B, and the reaction of A with B leads to the formation of C and D.

Let the reaction occur according to the following equation and form an equilibrium:

aA + bB cC + dD According to the law of mass action, at equilibrium, the rate of forward reaction

(r_f)=k_f [A]^a × [B]^b

Or rf = Kf

And that of backward reaction

(r_b)∝[D]^d × [E]e

Or,  r_b= k_b [D]^d× [E]^e

Where k and kb are the rate constants of the forward and the backward reactions respectively. (A), (B), (D), and (E) are the respective molar concentrations or active masses (mol.L^-1) of A, B, D, and E at equilibrium.

At equilibrium, the rate of the forward reaction (ry) = the rate of the backward reaction (r_b).

⇒ \(\text { So, } k_f[A]^a \times[B]^b=k_b[D]^d \times[E]^e\)

⇒ \(\text { or, } \frac{k_f}{k_b}=\frac{[D]^d \times[E]^e}{[A]^a \times[B]^b} \text { or, } \boldsymbol{K}=\frac{[\boldsymbol{D}]^d \times[E]^e}{[A]^a \times[B]^b}\)

The ratio of the two rate constants (fcy and kb) is a constant quantity at a given temperature.

So, K is a constant. The constant ‘K’ is called the equilibrium constant of the said reversible reaction. Equation (1) expresses the mathematical form of the law of mass action of the given reversible reaction.

In the expression of the equilibrium constant, the reaction. concentration terms of the reactants and the products represent their respective molar concentrations at equilibrium. They do not denote their initial concentrations

At constant temperature, the equilibrium constant (K) of a chemical reaction has a definite value. The value of K changes with temperature. This is because the changes of values of kJ- and kJ with the temperature change do not occur to the same extent.

Equilibrium constant

The equilibrium constant of a reaction is the ratio of the product of the active masses of products at equilibrium to the product of the active masses of reactants at equilibrium, with each active mass term raised to a power equal to its stoichiometric coefficient in the balanced equation of the reaction.

The equilibrium constant is also represented as K_c, K_p or K_x depending on whether the active mass is expressed in terms of molar concentrations partial pressure, or mole fraction.

Different kinds of equilibrium constants:

The law of chemical equilibrium:

This law states that when a reversible chemical reaction reaches equilibrium at a particular temperature, the ratio of the product of active masses of the products to that of the reactants, with each active mass term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation is constant.

The mathematical expression for the law of chemical equilibrium can be obtained by applying the law of mass action to a reversible reaction at equilibrium. For a general reversible reaction, aA + bB dD + cE; the mathematical expression can be written as, (constant at a particular temperature) [A]n[B];’

⇒ \(\frac{[D]^d[E]^e}{[\mathrm{~A}]^a[B]^b}=K\) = K (constant at a particular temperature

Expression of the equilibrium constant in case of a heterogeneous equilibrium

At a given temperature, the active mass or molar concentration of a pure solid or liquid is always taken as unity (1). For this reason, in the case of a heterogeneous equilibrium, the active mass or molar concentration term of a pure solid or liquid does not appear in the expression of the equilibrium constant.

Examples: The thermal decomposition of solid CaCO₃ in a closed container leads to the following equilibrium:

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

⇒\(K_c=\frac{[\mathrm{CaO}(\mathrm{s})]\left[\mathrm{CO}_2(\mathrm{~g})\right]}{\left[\mathrm{CaCO}_3(\mathrm{~s})\right]}=\left[\mathrm{CO}_2(\mathrm{~g})\right]\)

[CaCO₃(s)] = 1, [CaO(s)] = 1 ] and Kp = PCO₂(g)

As the value of Kc or Kp is constant of a given temperature, the molar concentration or the partial pressure of CO₂(g) at the equilibrium formed on the decomposition of solid CaC03 at a given temperature is always constant.

The following equilibrium is established during the vaporization of water in a closed vessel:

H₂O(l) ⇌ H₂O(g)

Here, the equilibrium constant

⇒  \(K_c=\frac{\left[\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right]}{\left[\mathrm{H}_2 \mathrm{O}(l)\right]}=\left[\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right]\)

Therefore, the molar concentration or the partial pressure of water vapor remaining in equilibrium with pure water at a particular temperature is always constant.

Expression of equilibrium constants of some chemical reactions:

Relation Between Different Equilibrium Constants

Relation between K_p and K_c

Let the following reversible gaseous reaction is at equiUbriumin a closed container at a certain temperature

aA(g) + bB(g);=± dD(g) + eE(g) If the molar concentrations of A(g), B(g), D(g) and E(g) at equilibrium be [A], [B], [D] and [£] respectively, and the partial pressures of A(g), B(g), D(g) and E(g) at equilibrium be pA, PiB pD and pp respectively, then

⇒ \(K_c=\frac{[D]^d \times[E]^G}{[A]^a \times[B]^b}\) ……………………(1)

⇒ \(K_p=\frac{p_D^d \times p_E^e}{p_A^a \times p_B^b}\) ……………………(2)

If the reaction mixture behaves as an ideal gas, then the ideal gas equation can be applied to each of the species present in the mixture.

If the pressure, temperature, and volume of n mole of an ideal gas are P, T, and V respectively,

Then \(P V=n R T \quad \text { or, } P=\left(\frac{n}{V}\right) R T=C R T\)

C= Molar concentration

Now by applying this equation to each species of the reaction mixture, we get

⇒ \(p_A=[A] R T, p_B=[B] R T, p_D=[D] R T \text { and } p_E=[E] R T \text {. }\)

Putting the values of pA, pB, PD, and pE into equation (2), we have

⇒ \(K_p=\frac{\{[D] R T\}^d \times\{[E] R T\}^e}{\{[A] R T\}^a \times\{[B] R T\}^b}\)

=\(\frac{[\mathrm{D}]^d \times[E]^e}{[A]^a \times[B]^b}(R T)^{(d+e)-(a+b)}\)

Or, \(K_p=K_c(R T)^{\Delta n}\)

Where, Δn = (d+ e)-(a + b) = the total number of moles of gaseous products – the total number of moles of gaseous reactants.

⇒ \(K_p=K_c(R T)^{\Delta n}\) ………………….(3)

Inequation (3), the value of Δn may be +ve, -ve, or zero. When the number of moles of the gaseous products is greater than, less than, or equal to the number of moles of the gaseous reactants, then the values of Δn become positive, negative, or zero, respectively.

If An is positive, K_p is greater than Kc. If Δn is negative, then the value of K_p is smaller than Kc. If Δn = 0, then K_p and K_c have the same value.

Relation between K_P and K_c in case of some chemical and physical changes:

Relation between K_P and K_x

Let, at a constant temperature, the following reversible gaseous reaction is at equilibrium in a closed vessel:

⇒ \(a A(g)+b B(g) \rightleftharpoons d D(g)+e E(g)\)

⇒ \(K_p=\frac{p_D^d \times p_E^e}{p_A^a \times p_B^b}\)……………….(1)

⇒ \(K_x=\frac{x_D^d \times x_E^e}{x_A^a \times x_B^b}\)……………….(2)

Where, P_A, P_B, P_D, and P_E are the partial pressures of A, B, D, and E respectively, at equilibrium, and xA, xB, xD and xE are their respective mole fractions at equilibrium. The partial pressure of a component gas in a gas mixture is its mole fraction times the total pressure of the mixture.

Hence, the relation between the partial pressures and mole fractions of the different components in the said gas mixture is

P_A = x_A xP, P_B = x_B xP, P_D = x_D xP, and P_E = x_E x P where P is the total pressure of the gas mixture at equilibrium.

Putting the values of P_A, P_B ,P_D, and P_E into equation (1), we have

⇒ \(K_p=\frac{\left(x_D \times P\right)^d \times\left(x_E \times P\right)^e}{\left(x_A \times P\right)^a \times\left(x_B \times P\right)^b}\)

= \(\frac{x_D^d \times x_E^e}{x_A^a \times x_B^b} \times(P)^{(d+e)-(a+b)}\)

= \(K_x \times(P)^{(d+e)-(a+b)}\)

∴ \(K_p=K_x \times P^{\Delta n}\) ……………….(3)

Here Δn = the total number of moles of gaseous products – the total number of moles of gaseous reactants. Equation (3) denotes the relation between Kp and Kx.

Relation between K_c and K_x

Let the following reversible gaseous reaction be at equilibrium at a temperature T and pressure P in a closed vessel

aA(g) + bB{g) ⇌  dD(g) + eE(g)

For this reaction, the relation between K_p and K_c is,

K_p = K_c (RT)^Δn………………………..(1)

And the relation between Kp and Kx is

, K_p = K_x (P)^Δn………………………..(2)

Where Δn = total number of moles of the gaseous products- total number of moles of the gaseous reactants. From equations (1) and (2), we have, Kc(RT)ÿn = Kx(P)ÿn

∴ , K_c = \(K_x\left(\frac{P}{R T}\right)^{\Delta n}\)………………………(3)

Equation (3) gives the relation between K_P and K_k.

Relation Between  K_P, K_c & K_x

K_P = K_c(RT)Δn = K_X(P)Δn When Δn = 0 for a reaction

Example: H₂(g) +I₂(g) ⇌ 2HI(g)

Or, N₂(g) + O₂(g) 2NO(g)], ⇌ then, K_P= K_c = K_x

Characteristics of the equilibrium constant

At constant temperature, the value of the equilibrium constant for each chemical reaction has a definite value. Temperature change brings about an increase or decrease in the value equilibrium constant.

• In the case of an endothermic reaction, the value of the equilibrium constant increases with the rise in temperature, while in the case of an exothermic reaction, the value of the equilibrium constant decreases with the rise in temperature
• The values of K_P and K_c are not influenced by pressure. If the temperature remains constant, the increase or decrease in pressure does not alter the values of K_P and K_P. However, except for a reaction for which Δn = 0, the value of depends upon pressure.
• The value of the equilibrium constant for any reaction neither increases nor decreases in the presence of a catalyst because the rate of both the forward and the backward reactions increase equally.
• The value of the equilibrium constant of a chemical reaction at a given temperature does not depend on the initial concentration of the reactants.

Example:

PCl₅(g)⇌ PCl₃(g) + Cl₂(g)

In this reaction, K_p at 450°C is 0.19. At 450°C, if the reaction is started with PCl₅(g) at any concentration, the value of K_p for the reaction is always 0.19.

The value of the equilibrium constant of any reaction depends on how the balanced equation for the reaction is written.

Consider the reaction in which NH₃(g) is synthesized from N₂ and H₂ gases. The balanced equation for this reaction is:

N₅(g) + 3H₂(g) ⇌ 2NH₃(g)

One can also write the equation as

⇒  \(\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g})\)

In case of(1), equilibrium constant

⇒ \(K_{c_1}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3},\)

While in case of (2), equilibrium constant, \(K_{c_2}=\frac{\left[\mathrm{NH}_3\right]}{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{H}_2\right]^{3 / 2}}\)

Thus, the values of Kcl and Kc are not the same. Comparingÿ andÿ gives =

⇒ \(K_{c_1}=K_{c_2}^2 \text {, i.e., } K_{c_2}=\sqrt{K_{c_1}}\)

Suppose, the equilibrium constant of the reaction,

aA + bB dD+eE is K.

NCERT Class 11 Chemistry Chapter 7 Equilibrium Important Topics

If the coefficients of the reactants and products are multiplied by then the equation becomes:

maA + mbB mdD + meE For this equation, the equilibrium constant, K = Km. Again, if the coefficients of the reactants and products are divided by m then the equation becomes:

⇒ \(\frac{1}{m} a A+\frac{1}{m} b B \rightleftharpoons \frac{1}{m} d D+\frac{1}{m} e E\)

For this equation, the equilibrium constant, K” = K^1/m

For any reversible chemical reaction, the values of ΔT equilibrium constants for the forward and the reverse reactions are reciprocal to each other.
Example:

⇒ \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\) for this reaction equilibrium constant

⇒ \(K_c=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right] \times\left[\mathrm{H}_2\right]^3}\)……………………(1)

If the reaction is started with NH₃ gas (product), then the reaction is:

2NH₃(g) ⇌ N₂(g) + 3H₂(g)

= \(\frac{\left[\mathrm{N}_2\right] \times\left[\mathrm{H}_2\right]^3}{\left[\mathrm{NH}_3\right]^2}\) ……………………(2)

From equations (1) and (2), we get \(K_c=\frac{1}{K_c{ }^{\prime}}\)

if a given reaction is expressed as the sum of two or more individual reactions, then the equilibrium constant of the given reaction equals the product of the equilibrium constants of the individual reactions.

If reaction (3) =reaction (2) + reaction (1), then equilibrium constant of reaction [3] = equilibrium constant of reaction [2] x equilibrium constant of reaction [1].

Example:

Reaction 1: \(A+B \rightleftharpoons C ; K_1=\frac{[C]}{[A][B]}\)

Reaction 2: \(C \rightleftharpoons D ; K_2=\frac{[D]}{[C]}\)

Reaction 3. \(A+B \rightleftharpoons D ; K_3=\frac{|D|}{[A][B]} .\)

Reaction 1+ reaction 2: \(A+B \rightleftharpoons D\)

∴ \(K_1 \times K_2=\frac{[C]}{[A][B]} \times \frac{[D]}{[C]}=\frac{[D]}{[A][B]}=K_3\)

Unit of the equilibrium constant

The unit of equilibrium constant depends on the difference between the sum of the exponents of the concentration (or partial pressure) terms in the numerator and that is the denominator of the equilibrium constant expression. Suppose, this difference is Δx. If Cl Δx = 0, then neither K_c nor K_p has a unit

Δx = 0, then both K_p and K_c have units.

K_p and K_c  for different values of Δx

Values of Δx and the unit of equilibrium constant for some reactions:

The equilibrium constant is unitless:

The term ‘active mass’ mentioned in the law of mass action is unitless. Consequently, Kp or Kc is also unitless. With the help of thermodynamics, it can be shown that the partial pressure of any species present in the expression of Kp is the ratio of measured pressure (P⁰) of that species at equilibrium to its standard pressure (P⁰). Since (P/P⁰) is unitless, Kp is also unitless.

In the case of a pure gas, standard pressure (P⁰) is taken as 1 atm. Similarly, the concentration of any species present in the expression of Kc is the ratio of the measured concentration (C) of that constituent at equilibrium to its standard concentration (C⁰). Since (C/C⁰) is unitless, Kc Is also unitless. The standard concentration (C⁰) of a solute dissolved in a solution is taken as 1(M) or 1 mol. L^-1

Graphical representations of some reversible reactions

1. Reaction: H₂(g)  +I₂ (g) ⇌ 2 HI(g)

Initialconcn.(mol.L^-1): 1.0 + 1.0 ⇌ 0

The equilibrium concentration of H₂(g) = 0.22mol.L^-1

2. Reaction:  2HI(g)+ ⇌ H₂(g)  +I₂ (g)

Initialconcn.(mol.L^-1): 2.0  ⇌ 0+  0

The equilibrium concentration of HI(g) = 1.56 mol. L^-1

3. Reaction:  N₂(g) +3H₂(g)  ⇌  2NH₃(g)

Initialconcn.(mol.L^-1): 1.0 + 3.0  ⇌ 0

The equilibrium concentration of NH₃(g) = 1.56 mol. L^-1

4. Reaction:  H₂(g) + CO₂(g)  ⇌ H₂O(g) + CO(g)

Initialconcn.(mol.L^-1): 2.0 + 1.0  ⇌ 0 + 0

The equilibrium concentration of H₂(g)= 1.6 mol. L^-1

Class 11 Chemistry Chapter 7 Equilibrium Summary

Numerical Examples

Question 1. At a particular temperature, the values of rate constants of forward and backward reactions are 1.5× 10^-2 L -mol^-1 .s^-1 and 1.8 × 10^-3 L-mol^-1.s^-1 respectively for the reaction A + B — C + D. Determine the equilibrium constant of the reaction at that temperature
Answer:

Equilibrium constant,

⇒ \(K=\frac{\text { Rate constant of forward reaction }}{\text { Rate constant of backward reaction }}\)

⇒ \(=\frac{1.5 \times 10^{-2} \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}}{1.8 \times 10^{-3} \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}}=8.33\)

Question 2. At a particular temperature, the equilibrium constant of the reaction 2A + B 2C is 8.0 × 10 L-mol^-1. If the rate constant of the reverse reaction is 1.24 L.mol^-1 .s^-1, then find the value of the rate constant of the forward reaction at that temperature.
Answer:

Equilibrium constant

⇒ \(K=\frac{\text { Rate constant of forward reaction }}{\text { Rate constant of reverse reaction }}\)

∴ The rate constant of the forward reaction = K x rate constant of the reverse reaction =

8 × 10⁴ ×1.24 L^-2.mol^-2.s^-1

= 9. 92 × 10⁴ L^-2.mol^-2.s^-1

Question 3. For the reaction 2SO₂(g) +O₂(g)⇌ 2SO₃(g), Kp = 3×1024 at 25°C. Find the value of ICc.
Answer:

For the given reaction, Δn = 2-(2 + 1) = -1

As given, Kp = 3 × 1024, T = (273 + 25) = 298K and

R = 0.0821 L-atm-mol^-1.K^-1

We know Kp = Kc(RT)n

Or, 3 × 10²⁴ = Kc(0.0821 × 298)^-1

K_c = 3 × 10²⁴ × 0.0821 × 298

= 7.34 × 10²⁵

Question 4. At 1500 K, Kc = 2.6 × 10-9 for the reaction 2BrFg(g) Br2(g) + 5F2(g). Determine the Kp of the reaction at that temperature.
Answer:

For the given reaction, Δn = (1+5)-2 = +4.

As per given data,

K_c = 2.6 × 10^-9

T = 1500K

Using R = 0.0821 L-atm-mol^-1.K^-1

In the equation Kp = Kc{RT)^Δn,

K_p = 2.6 × 10^-9× (0.0821 × 1500)4

= 0.598

Question 5. Find the temperature at which the numerical values of Kp and Kc will be equal to each other for the reaction \(\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g})\)
Answer:

In the case of the given reaction,

⇒ \(\Delta n=1-\left(\frac{1}{2}+\frac{3}{2}\right)=-1\).

If the numerical values of K_p and K_c be x, then for the above reaction, K_p = x atm^-1 and = x(mol. L^-1)-1=x L-mol^-1

Therefore, Kp = Kc(RT)^-1

Or, \(x \mathrm{~atm}^{-1}=x \mathrm{~L} \cdot \mathrm{mol}^{-1} \times \frac{1}{0.0821 \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times T}\)

∴ T=12.18K

∴ The numerical values of Kp and Kc for the given reaction will be equal to 12.18K.

Question 6. At 400°C, H₂(g) and I₂(g) are allowed to react in a closed vessel of 5 L capacity to produce in HI(g). At equilibrium, the mixture in the flask Is found to consist of 0.6 mol H₂(g), 0.6 mol I₂(g), and 3.5 mol I₂(g). Determine the value of Kc of the reaction.
Answer: Equilibrium of the reaction is:

H₂(g) + I₂(g)⇒2HI(g)

⇒ \(\text { Therefore, } K_c=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}\)

The capacity of the vessel = 5 L.

Hence, molar concentrations of H₂, I₂ and HI are:

⇒  \(\left[\mathrm{H}_2\right]=\frac{0.6}{5}=0.12 \mathrm{~mol} \cdot \mathrm{L}^{-1} ;\)

⇒   \(\left[\mathrm{I}_2\right]=\frac{0.6}{5}=0.12 \mathrm{~mol} \cdot \mathrm{L}^{-1} \text { and }[\mathrm{HI}]=\frac{3.5}{5}=0.7 \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

∴ \(K_c=\frac{(0.7)^2}{(0.12) \times(0.12)}=34.03\)

Question 7. At a particular temperature, CO(g) reacts with Cl₂(g) in a closed container to produce COCI₂(g). In the equilibrium mixture, partial pressures of CO(g), Cl₂(g), and COCl₂(g) are found to be 0.12, 1.2, and 0.58 atm respectively. Find the value of Kp of the reaction, \(\mathrm{CO}(g)+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}_2(\mathrm{~g}).\)
Answer:

For the given equilibrium \(K_p=\frac{p_{\mathrm{COCl}_2}}{p_{\mathrm{CO}} \times p_{\mathrm{Cl}_2}}\)

As given, P_CO =o 0.12 atm,P_Cl2= 1.2 atm, and

P_COCl = 0.58 atm at equilibrium.

∴ \(K_p=\frac{0.58}{0.12 \times 1.2}\)

= 4.03

Question 8. In a closed vessel of 1 dm₃ capacity, 1 mol N₂(g) and 2 mol H₂(g) interact to produce 0.8 mol NH₃(g) in the equilibrium mixture. What is the concentration of H₂(g) in the equilibrium mixture?
Answer:

Equation of the equilibrium reaction:

⇒ \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\)

It is observed from the reaction that 1 mol N₂(g) and 3 mol H₂(g) are necessary for the production of 2 mol NH₃(g).

Therefore, for the production of 0.8 mol NH₃(g), the number of moles of H2(g) required \(=\frac{3}{2} \times 0.8=1.2 \mathrm{~mol}\)

Hence, the number of moles of H2(g) remaining in the equilibrium mixture

= 2-1.2 = 0.8

And its molar concentration =0.8 mol.L^-1

Since 1 dm³=1L

Question 9. At 20°C, 0.258mol A(g) and 0.592 mol 5(g) are mixed in a closed vessel of capacity to conduct the following reaction: A(g) + 2B(g) C(g). If 0.035 mol C(g) remains in the equilibrium mixture, then determine the partial pressure of each constituent at equilibrium.
Answer:

According to the equation, 1 mol A(g) reacts with 2 mol 5(g) to produce 1 mol C(g).

Hence, 0.035 mol A(g) and 2 × 0.035 = 0.07 mol 5(g) are required to produce 0.035 mol C(g).

Therefore, equilibrium molar concentrations of different constituents will be as follows:

As given, T = (273 + 20) K = 293 K

∴ At equilibrium,

P_A = [A]BT =0.0446 × 0.0821 × 293 = 1.072 atm

P_B = [B]5T = 0.1044 × 0.0821 × 293 = 2.511 atm

P_C = [C]5T = 7 × 10^-3×  0.0821 × 293 = 0.168 atm

Equilibrium Chapter 7 NCERT Solutions for Class 11

Question 10. 2 mol of were heated in a sealed tube at 440°C until the equilibrium was reached. HI was found to be 22% dissociated. Calculate the equilibrium constant for the reaction

2HI(g) ⇌  H₂(g) +I₂(g)

Answer:

For the given reaction \(K_c=\frac{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}\)

As given in the question, HI(g) undergoes 22% dissociation. Hence, out of 2 mol HI(g),

2 × 0.22 = 0.44 mol HI(g) dissociates.

As obtained from the equation, 2 mol HI(g) dissociates to produce 1 mol H₂(g) and 1 mol I₂(g). Therefore, 0.44 mol HI(g) dissociates to form 0.22mol of each of H₂(g) and I₂(g).

If the volume of the container is V L, then the equilibrium molar concentrations of different constituents will be as follows:

= 1.56 /V

∴ \(K_c=\frac{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}=\frac{\frac{0.22}{V} \times \frac{0.22}{V}}{\left(\frac{1.56}{V}\right)^2}\)

= 0.0198

Question 11. 1 mol PCl₅(g) is heated in a closed container of 2-litre capacity. If at equilibrium, the quantity of PCl₅g) is 0.2 mol then calculate the value of the equilibrium constant for the given reaction

⇒ \(\mathrm{PCl}_5(g)\rightleftharpoons\mathrm{PCl}_3(g)+\mathrm{Cl}_2(g)\)

Answer:

For the above reaction \(K_c=\frac{\left[\mathrm{PCl}_3\right] \times\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}\)

As per the given data, the number of moles of PCl₅(g) at equilibrium =0.2. Hence, number of moles of PCl₅(g) dissociated =(1- 0.2)

= 0.8.

According to the equation, 1 mol PCl₅(g) dissociates to produce 1 mol of each of PCl₃(g) and Cl₂(g). Therefore, 0.8 mol PCl₅(g) will dissociate to give 0.8 mol of each ofPCl₃(g) and Cl₂(g).

As given, the volume of the container = 2 L. So, the equilibrium concentrations of different constituents are as follows:

∴ \(K_c=\frac{\left[\mathrm{PCl}_3\right] \times\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{0.4 \times 0.4}{0.1}\)

=1.6

Question 12. The following reaction is carried out at a particular temperature in a closed vessel of definite volume: CO₂(g) +H₂(g)s=± CO(g) + H₂O(g). Initially, partial pressures of CO₂(g) and H₂(g) are 2 atm and 1 atm respectively and that of CO₂(g) at equilibrium is 1.4 atm. Calculate the equilibrium constant of the reaction.

Answer:

For the given reaction \(K_p=\frac{p_{\mathrm{CO}} \times p_{\mathrm{H}_2 \mathrm{O}}}{p_{\mathrm{CO}_2} \times p_{\mathrm{H}_2}}\)

As given, partial pressure of CO₂(g) at equilibrium (Pco₂) = 1-4 atm. Hence, decrease in pressure of CO₂(g) until the equilibrium is reached =(2- 1.4)atm = 0.6atm According to the equation, 1 mol CO₂(g) reacts with 1 mol H₂(g) to form 1 mol CO(g) and 1 mol H₂O(g).

Therefore, if the pressure of CO₂(g) is reduced by 0.6 atm, then the pressure of H₂(g) will also be reduced by 0.6 atm and the pressure of each of CO(g) and H₂O(g) will be 0.6 atm Hence, partial pressures of different constituents at equilibrium will be as follows:

∴ \(K_p=\frac{p_{\mathrm{CO}} \times p_{\mathrm{H}_2 \mathrm{O}}}{p_{\mathrm{CO}_2} \times p_{\mathrm{H}_2}}\)

= \(\frac{0.6 \times 0.6}{1.4 \times 0.4}\)

= 0.64

Question 13. B(g) + C(g) ⇌  A(g). At constant temperature, a mixture of 1 mol A(g), 2 mol E(g), and 3 mol C(g) is left to stand in a closed vessel of 1 L capacity. The equilibrium mixture is found to contain B(g) of 0.175 molar concentration (mol.L^-1). Find the value of the equilibrium constant at that temperature.
Answer:

For the above reaction \(K_c=\frac{[A]}{[B] \times[C]}\)

As given, at equili-brium, [B] =0.175 mol.L^-1.

So, 2-0.175=1.825 mol.L^-1 of B participated in the reaction.

Now according to the equation, 1 mol B and 1 mol C combine to form 1 mol A. So, 1.825 mol of B and 1.825 mol of C combine to produce 1.825 mol of A. Thus at equilibrium, the concentration of A, B, and C will be

Since the volume of the container = 1L

∴ \(K_c=\frac{[A]}{[B] \times[C]}=\frac{2.825}{0.175 \times 1.175}=13.74\)

Reaction Quotient, Q

The reaction quotient of a reaction has the same form as the equilibrium constant expression. However, the values of molar concentrations (or partial pressures) in the expression of the reaction quotient are the values at any instant of the reaction, whereas these values in the equilibrium constant expression represent the equilibrium values.

Hence, at a particular temperature, the value of the equilibrium constant of a reaction is fixed but that of the reaction quotient is not.

Reaction Quotient:

At constant temperature, the reaction quotient of a reaction at any instant may be defined as the ratio of the product of molar concentrations (or partial pressures) of the products to that of the reactants, with each concentration (or partial pressure) term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation.

The reaction quotient is denoted by Q. When the reaction quotient is expressed in terms of the molar concentration of the reactants and the products, then it is represented by Qc.

The reaction quotient expressed by the partial pressures of the reactants and the products is represented by Q_P.

• At the start of the reaction, only reactants are present in the reaction system. So, the value of the numerator in the expression of Q (reaction quotient) is zero, and consequently Q = 0.
• If a reaction goes to completion, then only the products are present in the reaction system. So, the value of the denominator in the expression of Q is zero, and hence Q→∞.

If the reaction system contains both the reactants and products, then Q assumes a value in between zero and ∞.

1. For the reaction,  aA + bB ⇌  dD + eE,  the reaction quotient at any moment

 ⇒ \(Q_c=\frac{[D]^d \times[E]^e}{[A]^a \times[B]^b}\).

Where (A), (B), (D), and (E) are molar concentrations of A, B, D, and E respectively, at that moment.

For the same reaction, the expression of Kc:

 ⇒ \(K_c=\frac{[D]_{e q}^d \times[E]_{e q}^e}{[A]_{e q}^a \times[B]_{e q}^b}\)

Where [A]_eq, [B]_eq, [D]_eq, and [E]_eq are the molar concentrations of A, B, D, and E respectively, at equilibrium.

2. For the reaction, aA(g) + bB(g) ⇌  dD(g) + eE(g)

The reaction quotient at any instant,

⇒  \(Q_p=\frac{p_D^d \times p_E^e}{p_A^a \times p_B^b}\)

P_A, P_B, P_D, and P_E are the partial pressures of A, B, D, and E respectively, at that instant.

For the same reaction, the expression of Kp:

⇒  \(K_p=\frac{\left(p_D\right)_{e q}^d \times\left(p_E\right)_{e q}^e}{\left(p_A\right)_{e q}^a \times\left(p_B\right)_{e q}^b}\)

Where (P_A)_eq, (P_B)_eq, (P_C)_eq, and (Pp)eq are the partial pressures of, B, D, and E, respectively, at equilibrium.

Significance of reaction quotient:

At a particular temperature, by comparing the values of the reaction quotient (Q) of a chemical reaction at any instant and the equilibrium constant (K) of the reaction at that temperature, one can predict the extent to which the reaction has proceeded.

From the values of Q and AT, it is possible to predict whether the reaction has already reached or will reach the state of equilibrium. If equilibrium is not attained, then it can also be predicted whether the reaction will proceed in the forward or backward direction to achieve equilibrium.

Predicting the direction of reaction from the values of Q_c and K_P

Example: At 700 K, for the following reaction which is carried out in a closed vessel: H₂(g) + I₂(g) ⇌  2HI(g). Kc = 55.0 at 700K.

Analysis of the reaction mixture at a given moment during the reaction shows that the molar concentrations of H₂(g), I₂(g), and HI(g) are 1.8, 2.8, and 0.4 mol-L^-1, respectively. Is the reaction at equilibrium at that moment? If not, in which direction will we proceed to attain equilibrium?
Answer:

For the given reaction, \(Q_c=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}\)

At the moment of analysis \(Q_c=\frac{(0.4)^2}{1.8 \times 2.8}=0.0317\)

Hence, Q_c< K_c. Therefore, the reaction is not at equilibrium. To attain equilibrium, the value of Q_c will increase until it becomes equal to K_c.

Again, the value of Q_c will increase if, in its expression, the value of the Numerator increases and that of the denominator decreases. This is possible if the forward reaction occurs to a greater extent Therefore, the reaction will proceed more in the forward direction to attain equilibrium.

Applications Of Equilibrium Constant

Application-1:

At a given temperature, the value of the equilibrium constant indicates the extent to which a reaction has proceeded before it attains equilibrium.

Explanation:

In the expression equilibrium constant (K) of a reaction, the concentrations of the products appear in the numerator, while that of the reactants appears in the denominator.

Hence, a larger value of equilibrium constant (K) for any reversible reaction signifies higher concentrations of the products than reactants in the equilibrium mixture. This means reactants convert into products to a large extent before attaining equilibrium. Hence, the position of equilibrium lies far to the right.

Example:

Application 2: If the value of the equilibrium constant of a reaction at a given temperature is known, then the concentrations of the reactants and products at equilibrium can be calculated from the known initial concentrations of the reactants

Question 1. At 550 K, the value of the equilibrium constant (K£) is 0.08 for the given

PCl₅(g)⇌ PCl₃(g)+Cl₂(g) occurring in a closed container.

If the equilibrium concentration of PClg(g) and Cl₂(g) are 0.75 and 0.32 mol-L^-1 respectively, then find the concentration of PCl₃(g).
Answer:

For the given reaction \(K_c=\frac{\left[\mathrm{PCl}_3\right] \times\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}\)

As given, [PCl₅] = 0.75mol.L^-1

[Cl₅] = 0.32 mol.L^-1 and K_C = 0.08.

∴ \(\left[\mathrm{PCl}_3\right]=K_c \times \frac{\left[\mathrm{PCl}_5\right]}{\left[\mathrm{Cl}_2\right]}\)

= \(0.08 \times \frac{0.75}{0.32}=0.187 \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

∴ The equilibrium concentration of PCl₃ = 0.187mol.L^-1.

CBSE Class 11 Equilibrium Chapter 7 Key Points

Question 2. At a given temperature, Kp is 0.36 for the reaction, 2SO₂(g) + O₂(g)⇌ 2SO₃(g) occurring in a closed vessel. If at equilibrium, the partial pressures of SO₂(g) & O₂(g) are 0.15 atm & 0.8 atm respectively, then calculate the partial pressure of SO₃(g).
Answer:

For the given reaction, Kp \(K_p\)

= \(\frac{\left(p_{\mathrm{SO}_3}\right)^2}{\left(p_{\mathrm{SO}_2}\right)^2 \times p_{\mathrm{O}_2}}\)

As Given K_p =0.36, P_SO2= 0.15 atm and Po₂= 0.8 atm

∴ (P_SO3)2 = K_p ×(P_SO2×2 ×Po₂ = 036 × (0.15)2 × (0.8)

= 6.48 × 10^-3. atm or PSO₃= 0.08 atm

Question 3. For the reaction, A₂(g)+B₂(G) ⇌ 2AB (g), the value of the equilibrium constant is 50 at 100°C. If a flask of 1 L capacity containing 1 mol A₂ is connected with another flask of 2 L capacity containing 2 mol B₂, then calculate the number of moles of AB at equilibrium.
Answer:

Equilibrium constant, \(K_c=\frac{[\mathrm{AB}]^2}{\left[\mathrm{~A}_2\right] \times\left[\mathrm{B}_2\right]}\)

The equation of the reaction shows that the reduction of x mol of A₂ leads to the reduction of x mol of B₂ and the formation of 2x mol of AB. Hence, the number of moles of A₂, B₂, and AB at equilibrium will be as follows.

Now if the two flasks are connected, then the total volume of the reaction system becomes (1 + 2)L = 3L.

Hence, at equilibrium \(\left[\mathrm{A}_2\right]=\left(\frac{1-x}{3}\right) \mathrm{mol} \cdot \mathrm{L}^{-1}\)

⇒ \([B]=\left(\frac{2-x}{3}\right) \mathrm{mol}^{-1} \mathrm{~L}^{-1} \text { and }[\mathrm{AB}]=\left(\frac{2 x}{3}\right) \mathrm{mol} \cdot \mathrm{L}^{-1} \text {. }\)

Therefore,

⇒ \(K_c=\frac{\left(\frac{2 x}{3}\right)^2}{\left(\frac{1-x}{3}\right) \times\left(\frac{2-x}{3}\right)}=50 \text { or, } \frac{4 x^2}{(1-x)(2-x)}=50\)

or, 4x² =  50x²– 1 50x + 100

Or,  46x²– 150x+ 100 = 0

Or, \(x=\frac{150 \pm \sqrt{(150)^2-4 \times 46 \times 100}}{2 \times 46}=2.32 \text { or, } 0.934\)

∴ 2x = 4.64 or, 1.868

By the Initial number of moles of A₂ and B₂, the number of moles of AB cannot be 4.64. Therefore, the number of moles of AB at equilibrium = 1 .868.

Question 4. At a particular temperature, the value of Kp is 100 for the reaction, \(2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_2(g)+\mathrm{O}_2(g)\)occurringin a closed container. If the initial pressure of NO(g) is 25 atm then calculate the partial pressures of NO, N₂, and O₂ at equilibrium.
Answer:

For the given reaction \(K_p=\frac{p_{\mathrm{N}_2} \times p_{\mathrm{O}_2}}{\left(p_{\mathrm{NO}}\right)^2}\)

After the attainment of equilibrium, if the reduction in pressure of (g) is p atm, then partial pressures of different constituents at equilibrium will be as follows:

[The equation shows that at constant temperature and pressure if the pressure reduction of NO(g) is p atm then the increase in pressure of each of N₂(g) and O₂(g) will be P/2 atm.

⇒ \(\text { Hence, } K_p=\frac{\left(\frac{p}{2}\right) \times\left(\frac{p}{2}\right)}{(25-p)^2}=100 \text { or, }\left(\frac{p}{25-p}\right)^2=400\) \(\text { or, } \frac{p}{25-p}=20 \text { or, } 21 p=25 \times 20\) ∴ P= 23.8 atm

Therefore, at equilibrium, partial pressure of NO(g) = (25- 23.8) atm = 1.2 atm andpartial pressure of N₂(g) = partial pressure Of

⇒  \(\mathrm{O}_2(\mathrm{~g})=\frac{23.8}{2}=11.9 \mathrm{~atm}\)

Application-3: If the value of the equilibrium constant of any reaction at a constant temperature is known, then it is possible to predict whether the mixture of reactants and products is in equilibrium and if not, in which direction the reaction will proceed Formore discussion, to attains equilibrium article number at that 7.6. temperature.

Free Energy Change (ΔG) Of A Reaction and Equilibrium Constant

Suppose, at a temperature of TK, A and B react together to produce D and E according to the following equation: \(a A+b B \rightleftharpoons d D+e E\)

⇒ \(K=\frac{[D]_{e q}^d \times[E]_{e q}^e}{[A]_{e q}^a \times[B]_{e q}^b}[e q=\text { equilibrium }]\)…………………………..(1)

Where (A)_eq, (B)_eq,  (D) _eq, and (E )_eq are the molar concentrations of A, B, D, and E respectively, at equilibrium.

If ΔG is the from the energy of the system, then with the help of can show that

⇒ \(\Delta G=\Delta G^0+N T \ln \frac{[D]^a \times[B]^b}{[A]^a \times[B]^b}\)

or, ΔG = ΔG⁰+RT in Q…………………………..(2)

Where,

⇒ \(Q=\frac{[D]^d \times[B]^c}{[A]^a \times[B]^b}\) and (A),(B),(D ), and (E)

Represent the active masses or molar concentrations of A, B, I) and respectively at a given moment during the reaction. Is the reaction quotient.

ΔG° Is the standard free energy change of the reaction. If In a reaction, (lie molar concentration of each of the reactants and products. If unity, then the free energy change of the reaction is called (IK; standard free energy change (ΔG⁰).

For a reaction at constant temperature and pressure, if—

1. ΔG is negative, the reaction is spontaneous as written.
2. ΔG is positive, the reaction Is non-spontane O us as written.
3. ΔG is zero, the reaction Is at equilibrium.

Now, at the equilibrium of a reaction, ΔG = 0 and

⇒ \(Q=\frac{[D]_{e q}^a \times[B]_{e q}^e}{[A]_{e q}^a \times[B]_{e q}^b}=K \text { [equilibrium constant] }\)

Hence, from equation (2) we get, 0 = ΔG° + RTlnK or, ΔG° = -RTlnK …………………………..(3)

Or,ΔG° = -2.303RTlogK …………………………..(4)

Or, K= e^-ΔG0/Rt…………………………..(5)

Equations (3), (4), and (5) represent the relation between equilibrium constant {K) and the standard free energy change (ΔG°) of a reaction at a given temperature.

From equations (2) and (3), we get, ΔG = -RTlnK +RTlnQ

or ΔG = RT In \(\frac{Q}{K}\) …………………………..(6)

Equation (6) is called reaction isotherm.

If the equilibrium constant expression of a reaction is written in terms of the molar concentrations of reactants and products then K=Ke. This gives

⇒ \(\Delta G^0=-R T \ln K_c \text { or, } K_c=e^{-\Delta G^0 / R T}\)

For gaseous reactions, equilibrium constant expression is written in terms of the partial pressures of reactants and products. So, for such reactions K=K and AG° = -RTlnK p or, Kp = e-ΔG⁰/RT.

In the equation, reactant products taken the standards I molstatel, of 1, the equation the; -RT in K p, the standard state of each reactant and products Is considered to be 1 atm.

Significance Of the reaction ΔG°=-RT in K

1. At a particular temperature If the value of AG° is known, then the value of the equilibrium constant (K) can be calculated by using this equation. Similarly, If the equilibrium constant of a reaction at a given temperature is known, then the value of A <7° can be determined by using this reaction.
2. If ΔG° < 0, i.e., ΔG° = – ve, then K will be greater than [K> 1]. Under such conditions, the amount of products will be relatively more than that of the reactants present in the equilibrium mixture, i.e., the forward reaction predominates.
3. If ΔG° >0, i.e., ΔG = + ve, then K will be less than 1 [K < 1]. In such cases, the concentration of the reactants is greater than that of the products, i.e., the backward reaction predominates.

CBSE Class 11 Chemistry Notes For Free Energy Change (ΔG) Of A Reaction and Equilibrium Constant Numerical Examples

Question 1. For the reaction A(g) + 2B(g) ⇌ 2D(g), ΔG⁰– 2kJ.mol^-1 at 500 K. What is the value of Kp for the reaction ½A(g) + B(g) ⇌  D(g) at that temperature?
Answer:

From the equation ΔG⁰ = -RTlnKp, we get

⇒ \(\ln K_p=-\frac{\Delta G^0}{R T}=-\frac{2000 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{8.314 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1} \times 500 \mathrm{~K}}\)

=-0.4811

∴ K_p= 0.6181

∴ For the given reaction \(K_p=\frac{\left(p_D\right)^2}{p_A \times\left(p_B\right)^2}\)

If the equilibrium constant for the reaction

⇒ \(\frac{1}{2} A(g)+B(g) \rightleftharpoons D(g) \text { be } K_p^{\prime} \text {, then } K_p^{\prime}=\frac{p_D}{p_A^{1 / 2} \times p_B}\)

∴ \(K_p^{\prime}=\sqrt{K_p} \quad K_p^{\prime}=\sqrt{0.6181}=0.7862\).

Question 2. Find the value of ΔG⁰ and ICc for the following reaction at 298K. \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g)\) Given: standard free energy of formation (ΔG) of NO₂ and NO are 52.0 and 87.0 kj.mol^-1 respectively.
Answer:

ΔG° of a reaction = ∑ΔG_f⁰ of products -∑ΔG_f⁰ of reactants. For the given reaction,

⇒ \(\Delta G^0=\Delta G_f^0\left(\mathrm{NO}_2\right)-\Delta G_f^0(\mathrm{NO})-\frac{1}{2} \Delta G_f^0\left(\mathrm{O}_2\right)\)

= 52.0- 87.0- 0.0 = -35.0 kl -mol^-1

We know that, ΔG⁰=-RT In K_c

Or, -35 × 10³ J-mol^-1

= – 8.314 K^-1  mol^-1 ×  298K ln Kc

or; lnKc = 14.126

∴ Kc = 1.365 × 10⁶

CBSE Class 11 Equilibrium Chapter 7 Key Points

Question 3. At 298K, for the attainment of equilibrium of the reaction N₂O₄(g) ⇌ 2NO₂5mol of each of the constituents is taken. Due to this, the total pressure of the mixture turns 20 atm. If ΔG_f⁰ (N₂O₄)= 100 kJ-mol^-1 and ΔG_f⁰ (NO₂) = 50 k(J-moI_1) then

1. Give the value of ΔG of the reaction.
2. In which direction will the reaction proceed to equilibrium?

Answer:

For the reaction,

⇒ \(\Delta G^0 & =2 \Delta G_f^0\left(\mathrm{NO}_2\right)-\Delta G_f^0\left(\mathrm{~N}_2 \mathrm{O}_4\right)\)

= \((2 \times 50-100)=0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Total number of moles in the reaction mixture =5 + 5 =10

⇒ \(\text { So, } p_{\mathrm{N}_2 \mathrm{O}_4}=\frac{5}{10} \times 20=10 \mathrm{~atm} \& p_{\mathrm{NO}_2}\)

= \(\frac{5}{10} \times 20=10 \mathrm{~atm}\)

Therefore, the Q_p of the reaction

⇒ \(\frac{\left(p_{\mathrm{NO}_2}\right)^2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{(10)^2}{10}=10\)

1. We know, ΔG = ΔG⁰ + RT]nQp

∴ ΔG =0 + 8.314 ×  298 In 10 = 5.706 kJ

Since ΔG⁰ =0

Again, ΔG⁰ =-RT In Kp;

So, 0 = -RT In Kp [since ΔG⁰ = 0]

or, Kp = 1

Since, Qp > Kp, the reaction wall proceeds more toward the left to attain equilibrium.

CBSE Class 11 Chemistry Notes For Determination Of Equilibrium Constants Of Some Reactions

Esterification of alcohol:

CH₃COOH(I) + C₂H₅OH(I) ⇌  CH₃COOC₂H₅(Z) + H₂O(Z)

Let at a particular temperature, a mol of acetic acid (CH₃COOH) reacts with b mol of ethyl alcohol (C₂H₅OH) to produce x mol of ethyl acetate (CH₃COOC₂H₅) and x mol of water (H₂O) at equilibrium.

According to the balanced equation, for the formation of x mol of ester and x mol of H_2O, x mol of CH₃COOH and x mol of C₂H₅OH are required. If the volume of the reaction mixture is by V L, the die concentration of the different species at equilibrium as well as

The expression of the equilibrium constant will be as given in the following table:

Formation of NO(g) from N₂(g) and O₂(g): For the reaction:

⇒ \(\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\);

Let us assume, at a constant temperature a mol of N₂ reacts with b mol of O₂ in a closed container of VL to produce 2xmol of NO at equilibrium.

According to the equation,ifx mol of N₂ and O₂ reacts with each other, then 2x mol of NO are formed.

Let, the total pressure of the reaction mixture at equilibrium = P. The molar concentrations and partial pressures of the reactants and products at equilibrium and

The expressions of K_p and K_c are given in the table:

Dissociation of PCl₅ gas: For the reaction:

PCl₅(g)⇌PCl₃(g) + Cl₂(g);

Let us assume, at a particular temperature, 1 mol of PCl₅ gas is heated in a closed vessel of V L capacity so that x mol of PCI5 gets dissociated at equilibrium.

According to the given equation, x mol of PCl₅ on dissociation produces x mol of PCl₃ and x mol of Cl₂.

1. Let us assume that die total pressure of the reaction mixture at equilibrium is P. Hence, the molar concentrations and partial pressures of the reactant and products at equilibrium

The expression equilibrium constants are given in the following table:

Formation of NH₃ from N₂ and H₂:

For the reaction: N₂(g) + 3H₂(g) v=± 2NH₃(g);

Let us assume, at a particular temperature, 1 mol of N₂(g) is allowed to react with 3 mol of H₂(g) in a closed vessel of V L capacity. After the attainment of equilibrium, if 2x mol of NH₃(g) is produced, then according to the equation x mol of N₂(g) and 3x mol of H₂(g) would be consumed.

Let, the total pressure of the reaction mixture be at equilibrium. Therefore, the molar concentrations and partial pressures of the reactants and product at equilibrium and the expression of equilibrium constants are given in the following table

Thermal decomposition of ammonium carbamate:

Reaction:

⇒\(\mathrm{NH}_2 \mathrm{CO}_2 \mathrm{NH}_4(\mathrm{~s}) \Rightarrow 2 \mathrm{NH}_3(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})\).

Let us assume, the l mol of NH₂CO₂NH₄(s) is heated at a particular temperature in a closed vessel of volume VL.

At equilibrium, if x mol of NH₂CO₂NH₄(g) undergoes decomposition, then, according to the equation, 2x mol of NH₃ and x mol of CO₂, will be produced.

The total pressure of the reaction mixture at equilibrium = P, then the molar concentrations and partial pressures of products and the expression of equilibrium constants are given in the table:

CBSE Class 11 Chemistry Notes For Relation Between Degree Of Dissociation And Degree Of Association With Vapour Density

When a compound undergoes incomplete dissociation, an equilibrium is established between the undissociated molecules of the die compound with the species formed on dissociation.

The extent of dissociation of a compound is usually quantized in terms of the degree of dissociation. It is defined as the ratio of the number of moles of the compound that dissociates to the initial number of moles of the compound.

Similarly, if a compound undergoes association, its extent of association is quantized in terms of degree of association. It is defined as the ratio of the number of moles of die compound associates to the initial number of moles of the compound.

Relation between the vapor density and degree of dissociation

Suppose, each molecule of gas A₂ on dissociation forms n molecules of gas B. If the reaction is initiated with 1 mol of gas A in a closed vessel of volume V at constant temperature and pressure, then the die number of moles of A₂ and B at equilibrium are as follows-

Where = the degree of dissociation of A₂

∴Total number of moles of the mixture at equilibrium =1- α + nα =1 + α(n- 1) mol

At constant temperature and pressure, the total volume of [1 + o(n- 1)] mol of the gas mixture is [1 + α(n- 1)× V].

Let the actual density and vapor density of gas A₂ before dissociation be d and D, respectively. Since the volume of the system increases on the dissociation of gas A₂ the density and vapour density of the gas mixture at equilibrium will be different from the actual values of these two quantities for gas A₂.

Suppose, the observed density and vapor density of the gas mixture at equilibrium be d’ and D’, respectively.

Since the mass of the system remains the same, we can write

d ×  V = d’ ×  V[1 + α(n- 1)]

Or, \(\frac{d}{d^{\prime}}=1+\alpha(n-1)\)……………………………….(1)

\(\frac{D}{D!}=1+\alpha(n-1)\)……………………………….(1)

∴ \(\frac{d}{d^{\prime}}=\frac{D}{D^{\prime}}\)

From equation (1) we get \(\alpha=\frac{d-d^{\prime}}{d^{\prime}(n-1)}\)

From equation (2) we get \(\alpha=\frac{d-d^{\prime}}{d^{\prime}(n-1)}\)

Relation between the vapor density and degree of association

Let us consider n mol of gas A associated to form 1 mol of gas B. If the reaction is initiated with 1 mol of gas A in a closed vessel of volume V at constant temperature and pressure,

Then the number of moles of A and B at equilibrium are as follows:

where a = the degree of association of A

∴ Total number of moles of the mixture at equilibrium

⇒ \(=1-\alpha+\frac{\alpha}{n}=1-\alpha\left(1-\frac{1}{n}\right) \mathrm{mol}\)

At constant temperature and pressure, the total volume of

⇒ \(\left[1-\alpha\left(1-\frac{1}{n}\right)\right]\) mol gas

⇒ \(\left[1-\alpha\left(1-\frac{1}{n}\right) \times v\right]\)

Now, let actual density and vapor density of gas A before association be d and D respectively.

Chemical Equilibrium Class 11 NCERT Solutions

If the observed density and vapor density of the gas mixture at equilibrium are d’ and D’, respectively, then—

⇒ \(d \times V=d^{\prime} \times V\left[1-\alpha\left(1-\frac{1}{n}\right)\right]\)

Or,  \(\frac{d}{d^{\prime}}=1-\alpha\left(1-\frac{1}{n}\right)\)…………………………(1)

⇒ \(\frac{D}{D^{\prime}}=1-\alpha\left(1-\frac{1}{n}\right)\) …………………………(2)

Since ⇒ \(\frac{d}{d^{\prime}}=\frac{D}{D^{\prime}}\)

From equation (1) we get

⇒ \(\alpha=\frac{d^{\prime}-d}{d^{\prime}\left(1-\frac{1}{n}\right)}\)

From equation (2) we get

⇒ \(\alpha=\frac{D^{\prime}-D}{D^{\prime}\left(1-\frac{1}{n}\right)}\)

CBSE Class 11 Chemistry Notes For Chapter 7 Equilibrium Ionic Equilibrium Introduction

The compounds that conduct electricity in a molten state or solution and dissociate chemically into new substances are called electrolytes.

Various acids Example: HCl, HNO₃, H₂SO₄),

Bases Example: NaOH, KOH)

Salts Example: NaCl, KCl, CuSO₄, AgNO₃)

Dissolve in water to conduct electricity. Hence, these are electrolytes.

On the both er hand, substances that are unable to conduct electricity either in a molten state or in solution are known as non-electrolytes. Glucose, sugar, alcohol, benzene, etc., are examples of non-electrolytes. In an aqueous solution (or in a molten state), electrolytes undergo spontaneous dissociation or ionization to produce positively and negatively charged particles or ions.

This is known as electrolytic dissociation or ionization. The ions can conduct electricity in the solution. In a particular solvent and at a certain temperature and concentration, the degree of dissociation or ionization of any electrolytic substance depends on the nature of that substance. The fraction of the total amount of dissolved electrolyte that exists in a dissociated or ionized state is called the degree of dissociation or ionization.

Depending on the degree of ionization in an aqueous solution, electrolytes can be classified into two categories Strong electrolytes and Weak electrolytes. The electrolytes which dissociate almost completely in aqueous solution at all concentrations are called strong electrolytes.

Strong acids Example: HCl, HNO₃, H₂SO₄ ),

Strong bases Example: NaOH, KOH

Most of the salts Example: NaCl, KC1, NH₄Cl, CuSO₄ )

Are strong electrolytes. Electrolytes that dissociate partially aqueous solution are called weak electrolytes.

Some salts Example: BaSO₄, HgCl₂

Most of the organic acids Example: HCOOH, CHgCOOH

A few in organic bases Example: Fe(OH)₃, NH₃ ),

Some inorganic acids Example: HCN, and H₂CO₃ ) are weak electrolytes. In aqueous solutions, strong electrolytes undergo almost complete ionization.

Hence, such ionisations are represented by a single arrow(→)

For example:

NaCl(aq)→ Na⁺(aq) + Cl^–(aq); HCl(aq)→H⁺(aq) + Cl^– (aq).

On the other hand, weak electrolytes undergo partial ionization in aqueous solution. Hence, such ionisations are reversible. Consequently, in an aqueous solution of weak electrolytes, a dynamic equilibrium exists between the dissociated ions and unionized molecules. This is known as ionic equilibrium.

Due to its irreversible nature, such ionisations are represented by double arrows(→).

For example:

HCN(aq) ⇌  H+(ag) + CN^–(aq); CH₃COOH(aq) ⇌  CH₃COO^–(aq) + H⁺(aq)

NCERT Solutions For Class 11 Chemistry Chapter 2 Structure Of Atom Multiple Choice Questions

Question 1. The ionization potential of a hydrogen atom is 13.6 eV. A hydrogen atom in the ground state is excited by monochromatic light of energy 12.1 eV. The spectral lines emitted by hydrogen according to Bohr’s theory will be.

1. One
2. Two
3. Three
4. Four

Answer: 3. Three

Question 2. The number of waves made by a Bohr electron in an orbit of maximum magnetic quantum number 3 is-

1. 4
2. 3
3. 2
4. 1

Answer: 1. 4

Question 3. In which of the following cases would the probability of finding an electron residing in a d -d-orbital be zero

1. xz and yz -planes
2. xy and yz -planes
3. z -direction, yz and xz -plane
4. xy and xz -planes

Answer: 1. xz and yz -planes

Structure of Atom Class 11 Multiple Choice Questions

Question 4. An electron beam with a de Broglie wavelength of P A is accelerated till its wavelength is halved. By what factor will kinetic energy change

1. 2
2. 1/4
3. 4
4. none

Answer: 3. 1/4

Question 5. If the Aufbau rule is not followed then the percent change in total (n + l) value for unpaired electrons in 25Mn is-

1. 60
2. 50
3. 40
4. 30

Answer: 3. 40

Question 6. If the shortest wavelength of the IT -atom in the Lyman series is X, then the longest wavelength in the Paschen series of He+ is

1. \(\frac{36 X}{5}\)
2. \(\frac{36 X}{7}\)
3. \(\frac{7 X}{36}\)
4. \(\frac{6 X}{5}\)

Answer: 2. \(\frac{36 X}{7}\)

Question 7. The atomic numbers of elements X, Y, and Z are 19, 21, and 25 respectively. The number of electrons present in the ‘M’ shells of these elements follows the order —

1. Z<Y<x
2. X<y<Z
3. Z>X>Y
4. Y>Z>X

Answer: 1. Z<Y<x

Question 8. Hydrogen atoms are excited in the n = 4 state. In the spectrum of the emitted radiation, the number of lines in the ultraviolet and visible regions are respectively—

1. 2:3
2. 3:1
3. 1:3
4. 3:2

Answer: 4. 3:2

Question 9. Which orbital has only a positive value of wave function at all distances from the nucleus—

1. 3d
2. 2p
3. 2s
4. 1s

Answer: 4. 1s

Question 10. The number of photons of light having wave number ‘a’ in the 32 energy source is

1. \(\frac{h c}{3 a}\)
2. 3hca
3. \(\frac{3}{h c a}\)
4. \(\frac{3}{h c a}\)

Answer: 3. \(\frac{3}{h c a}\)

Question 11. The wavelength of the de Broglie wave of the electron in the sixth orbit of-atom is—( r₀ = Bohr’s radius)

1. πr₀
2. 12πr₀
3. 6πr₀
4. 24πr₀

Answer: 2. 12πr₀

Question 12. In an orbit, the velocity of an electron in the excited state of Hatom is 1.093 × 108 cm^-1 The circumference of this orbit is 

1. 13.3Å
2. 6.65Å
3. 3.33Å
4. 26.65Å

Answer: 1. 13.3Å

Question 13. Which have the largest number of unpaired electrons in p -p-orbitals in their ground state electronic configurations—

1. Te, I, Xe
2. F, Cl, Br
3. Ne, Ar, K
4. N, P, As

Answer: 4. N, P, As

NCERT Solutions Class 11 Chemistry Chapter 2 MCQs

Question 14. Which orbitals have two nodal planes passing through the nucleus —

1. d
2. p
3. s
4. None

Answer: 1. d

Question 15. Compared to the mass of the lightest nuclei, the mass of an electron is only-

1. 1/80
2. 1/800
3. 1/1800
4. 1/2800

Answer: 3. 1/1800

Question 16. Among the following sets of quantum numbers, which one Is Incorrect for 4d -electrons-

1. \(4,3,2,+\frac{1}{2}\)
2. \(4,3,2,+\frac{1}{2}\)
3. \(4,2,-2, \frac{1}{2}\)
4. \(4,2,1, \frac{-1}{2}\)

Answer: 2. \(4,3,2,+\frac{1}{2}\)

Question 17. Which d -orbitals have a different shape from the rest of all d orbitals—

1. \(d_{x^2-y}\)
2. dx
3. dz²
4. d_yz

Answer: 3. dz²

Question 18. Which element possesses non-spherical shells

1. He
2. B
3. Be
4. Li

Answer: 2. B

Question 19. Which have the same number of s -electrons as the d electrons In Fe2+

1. Li
2. Na
3. Na
4. Fe

Answer: 4. Fe

Class 11 Chemistry Chapter 2 Atom Structure MCQs

Question 20. An anion X³ has 36 electrons and 45 neutrons. What is the mass number ofthe element X-

1. 81
2. 84
3. 78
4. 88

Answer: 3. 78

Question 21. Consider the set of quantum numbers \(3,2,-2,+\frac{1}{2}\), if the given subshell is filled. The next electron will enter the orbital with n and l value

1. n = 3, l  = 3
2. n = 4, l = 1
3. n =1,l = 1
4. n = 2, l-1

Answer: 2. n=4,l=1

Class 11 Chemistry Chapter 2 Atom Structure MCQs

Question 22. Given that an orbital is symmetric about the nucleus, then the value of azimuthal quantum number and magnetic quantum number are respectively

1. -1+1
2. +1+1
3. 0,0
4. 1,0

Answer: 3. 0,0

Question 23. A certain F.M. station broadcasts at a wavelength equal to 3.5 m. How many photons per second correspond to the transmission of one kilowatt–

1. 2.24×10²⁷
2. 1.76×10²⁸
3. 2.26×10²⁸
4. 1.43×10²⁶

Answer: 2. 1.7 6 ×10²⁸

Question 24. A Bohr orbit in H-atom has a radius of 8.464 A. How many transitions may occur from this orbit to the ground state-

1. 10
2. 3
3. 6
4. 15

Answer: 3. 6

Question 25. The angular momentum of the electron in the 4/-orbital of a one-electron species according to wave mechanics is —

1. \(\sqrt{3} \frac{h}{\pi}\)
2. \(2 \frac{h}{\pi}\)
3. \(\sqrt{\frac{3}{2}} \frac{h}{\pi}\)
4. \(\sqrt{\frac{1}{2}} \frac{h}{\pi}\)

Answer: 1. \(\sqrt{3} \frac{h}{\pi}\)

Question 26. Consider the Structure of the ground Atom Mato of cnCr atom (z= 24).-The1 number of electrons with the azimuthal quantum numbers, =1 and 2 nrc respectively

1. 12 and 4
2. 12 and 5
3. 16 and 4
4. 16 and 5

Answer: 2. 12 and 5

Class 11 Chemistry Chapter 2 Atom Structure MCQs

Question 27. The magnetic moment of Mx+ (atomic number of M = 25 ) is Jl5 BM. The number of unpaired electrons and the value of x respectively are—

1. 5,2
2. 3,2
3. 3,4
4. 4,3

Answer: 3. 3,4

Question 28. Radial part of the wave function depends upon quantum numbers

1. n and s
2. 1 and m
3. 1 and s
4. n and 1

Answer: 1. n and s

Question 29. Which ofthe following pairs of nuclides are in diapers

1. \({ }_6^{13} \mathrm{C} \text { and }{ }_8^{16} \mathrm{O}\)
2. \({ }_6^{13} \mathrm{C} \text { and }{ }_8^{16} \mathrm{O}\)
3. \({ }_1^3 \mathrm{H} \text { and }{ }_2^4 \mathrm{He}\)
4. \({ }_{25}^{55} \mathrm{Mn} \text { and }{ }_{30}^{65} \mathrm{Zn}\)

Answer: 4. \({ }_{25}^{55} \mathrm{Mn} \text { and }{ }_{30}^{65} \mathrm{Zn}\)

Question 30. The dissociation energy of H2 is 430.53 kfrmol-1. If hydrogen is dissociated by illumination with radiation of wavelength 253.7 nm, the fraction of the radiant energy that will be converted into kinetic energy is given by

1. 100%
2. 8.76%
3. 2.22%
4. 1.22%

Answer: 2. 8.76%

Question 31. The correct order of penetrating power of 3s, 3p, and 3d electrons is

1. 3d>3p>3s
2. 3s>3p>3d
3. 3s>3p>3d
4. 3d>3s>3p

Answer: 3. 3s>3p>3d

Question 32. Hund’s rule pertains to the distribution of electrons in

1. Principal energy shell
2. An orbital
3. Degenerate
4. None of these

Answer: 3. Degenerate

Question 33. A principal shell having the highest energy subshell to be V can accommodate electrons to a maximum of—

1. 18
2. 32
3. 25
4. 50

Answer: 4. 50

NCERT Class 11 Structure of Atom MCQs

Question 34. When an electron of H-atom jumps from a higher to lower energy, then—

1. Its potential energy increases
2. Its kinetic energy increases
3. Its angular momentum remains unchanged
4. Its de Broglie wavelength increases

Answer: 4. Its de Broglie wavelength increases

Question 35. What will be the number of spectral lines (AO observed if an electron undergoes a transition from n2 excited level to nl excited level in an atom of hydrogen—

1. \(N=\frac{\left(n_2-n_1\right)\left(n_2-n_1+1\right)}{2}\)
2. \(N=\frac{\left(n_1-n_2\right)\left(n_2-n_1+1\right)}{2}\)
3. \(N=\frac{\left(n_2+n_1\right)\left(n_1+n_2+1\right)}{2}\)
4. N = 2(nl-n2)(n2 + n1-l)

Answer: 1. \(N=\frac{\left(n_2-n_1\right)\left(n_2-n_1+1\right)}{2}\)

NCERT Solutions Chapter 2 Structure of Atom MCQs

Question 36. The given diagram indicates the energy levels of certain atoms. When the system moves from 2E level to E, a photon of wavelength X is emitted. The wavelength of photon produced during its transition from level to E iS-

1. \(\frac{\lambda}{3}\)
2. \(\frac{3 \lambda}{3}\)
3. \(\frac{4}{3} \lambda\)
4. 3 λ

Answer: 4. 3 λ

Question 37. Electromagnetic radiation with maximum wavelength Is-

1. Ultraviolet
2. Radiowaves
3. X-way
4. infrared

Answer: 2. Radiowaves

Question 38. Brackett series are produced when the electrons from the outer orbits jump to—

1. 2nd orbit
2. 3rd orbit
3. 4th orbit
4. 5th orbit

Answer: 3. 4th orbit

Question 39. The following sets that do NOT contain isoelectronic species—

1. \(\mathrm{BO}_3^{3-}, \mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\)
2. \(\mathrm{SO}_3^{2-}, \mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\)
3. \(\mathrm{CN}^{-}, \mathrm{N}_2, \mathrm{C}_2^{2-}\)
4. \(\mathrm{PO}_4^{3-}, \mathrm{SO}_3^{2-}, \mathrm{ClO}_4^{-}\)

Answer: 2. \(\mathrm{SO}_3^{2-}, \mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\)

Atom Structure MCQs Class 11 NCERT

Question 40. The hydrogen-like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light, the ion undergoes a transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy ofthe hydrogen atom-

The state is

1. 1s
2. 2s
3. 2p
4. 3s

Answer: 2. 2s

NCERT Solutions For Class 11 Chemistry Chapter 3 Classification Of Elements And Periodicity In Properties Multiple Choice Questions

Question 1. An element belongs to Group 15 and the third period of the periodic table. Its electronic configuration will be

1. ls²2s²2p²
2. ls²2s²2p⁴
3. ls²2s²2p⁶3s²3p³
4. ls²2s²2p⁶3s²3p²

Answer: 3. ls²2s²2p⁶3s²3p³

The electronic configuration of the valence shell of group 15 elements in the periodic table is ns²np³ where n is the tire period number.

Therefore, the element located in the third period has the electronic configuration ls²2s²2p⁶3s²3p³

Questions 2. Which one ofthe following has the lowest ionization energy

1. ls²2s²2p⁶
2. ls²2s²2p⁵
3. ls²2s²2p⁶3s¹
4. ls²2s²2p³

Answer: 2. ls²2s²2p⁵

The electronic configuration ls²2s²2p⁶3s¹ is that of an alkali metal. In a certain period of the periodic table, the ionization potential of alkali metals is the lowest.

Question 3. If 2nd the ionization 1st ionization energy of the atom-atom is — is 13.6 eV, then the

1. 27.2 eV
2. 40.8 eV
3. 54.4 eV
4. 108.8eV

Answer: 3. 54.4 eV

The first ionization energy of the -atom is calculated as:

⇒ \(\frac{2 \pi^2 m Z^2 e^4}{h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=13.6 \mathrm{eV}\)

The second ionization energy of the He atom is = 13.6 X Z2 eV = 13.6 X 22 eV = 54.4 eV.

Classification of Elements and Periodicity in Properties MCQs Class 11

Questions 4. The stable bivalency of Pb and trivalency of Bi is

1. Due to d contraction in pb and bi
2. Due to relativistic contraction of the 6s -orbitals of pb
3. And bi, leading to an inert pair effect
4. Due to the screening effect
5. Due to the attainment of noble gas configuration.

Answer: 2. Due to relativistic contraction of the 6s -orbitals of pb

Due to the relativistic contraction of 6s -orbital, an inert pair effect is observed as a result of which, the lower oxidation states of elements attain stability on moving down a group in the periodic table. Thus lead shows stable bivalency while bismuth shows stable trivalency

Question 5. Which of the following is correct?

1. Radius of ca²⁺ < cl- < s²
2. The radius of cl- <s^2-ca²⁺
3. Radius of s^2- = cl- = ca²⁺
4. Radius of s^2- cl- < ca²⁺

Answer: 1. Radius of ca²⁺ < cl- < s²⁺

The number of electrons in Ca²⁺, S^2- and CI^–= 18, i.e., all three ions are isoelectronic.

For any ion, ionic radius oc(e/Z) [where, e = electron number and Z = atomic number]. Therefore, the correct order of the radii ofthe given three ions is—Ca²⁺ < Cl^– < S^2-

Question 6. For BCl₃, AlCl₃, and GaCl₃, the increasing order of ionic character is

1. BCl₃ < AlCl₃ < GaCl₃
2. GaCl₃ < AlCl₃ < BCl₃
3. BCl₃ <GaCl₃ <AlCl₃
4. AlCl₃ < BCl₃ < GaCl₃

Answer: 3. BCl₃ <GaCl₃ <AlCl₃

The ionic character of a compound depends on the polarising power of the cation present in the compound. The greater the magnitude of polarising power, the lesser the ionic character of the cation.

Polarisingpower of B³⁺, Al³⁺, and Ga³⁺ follows the order: Al³⁺ < Ga³⁺ < B³⁺, Thus, the order of increasing ionic character is— BCl₃ < GaCl₃ < AICI₃.

Question 7. The hydrides of the first elements in groups 15-17, namely NH₃, H₃O, and HF respectively show abnormally high values for ting and boiling points. This is due to

1. Small sizes of n, 0 and
2. The ability to form extensive intermolecular-bonding
3. The ability to form extensive intramolecular h-bonding
4. Effective van der waals interaction

Answer: 2. The ability to form extensive intramolecular-bonding

The ionization potential values of, O, and F are quite high and thus, they form intermolecular H-bonding in the compounds, NH₃, H₃O, and HF. As a result, these compounds show abnormally high values for melting and boiling points.

Question 8. Decreasing the basic character of K₂O, BaO, CaO, and MgO is

1. K₂O > BaO > CaO > MgO
2. K₂O > CaO > BaO > MgO
3. MgO > BaO > CaO > K₂O
4. MgO > CaO > BaO > K₂O

Answer: 1. K₂O > BaO > CaO > MgO Alkali metal oxides are highly basic. Down the group in the periodic table, the ionization potential of the alkaline earth metals decreases.

Question 9. Amongst Be, B, Mg, and A₁ the second ionization potential is maximum for

1. B
2. Be
3. Mg
4. Al

Answer: 1. B

The electronic configuration of B (5) is ls²2s²2p¹. The second ionization potential of B is the maximum because an electron has to be removed from the filled 2s orbital which will require a high amount of energy.

Question 10. An element X belongs to the period and fifteenth group of the periodic table. Which of the following statements is true

1. It has a filled s -s-orbital and a partially filled d -d-orbital
2. It has filled s -and p -orbital and a partly filled d -orbital
3. It has filled s -and p orbitals and a half-filled d -orbital.
4. It has a half-filled p -p-orbital and filled s and d – d-orbital

Answer: 4. It has a half-filled p-orbital and filled s and d – orbital

The element X is positioned at the 4th period and 15th group. Hence the element is As. (Atomic mass = 33) Electronic configuration ofthe element: ls²2s²2p⁶3s²3p⁶3d¹⁰4s²43³ Thus the element has a half-filled p-orbital and a filled s  -and d – orbital

Question 11. Which of the following atoms should have the highest 1st electron affinity—

1. F
2. O
3. N
4. C

Answer: 1. F

Decreasing order of electron affinity: F > O > N > C.

Question 12. Which of the set of oxides are arranged in the proper order of basic, amphoteric acidic

1. SO₂,P₂O₅,CO
2. BaO, Al₂O₃, SO₂
3. CaO, SiO₂> Al₂O₃
4. CO₂, Al₂O₃, CO

Answer: 2. BaO→basic, Al₂O₃ amphoteric, SO₂ → acidic

Question 13. Which ofthe following orders presents a correct sequence of the increasing basic nature ofthe given oxides—

1. Al₂O₃ < MgO < Na₂O < K₂O
2. MgO < K₂O < Al₂O₃ < Na₂O
3. Na₂O < K₂O < MgO < Al₂O₃
4. K₂O < Na₂O < Al₂O₃ < MgO

Answer: 1. Al₂O₃ < MgO < Na₂O < K₂O

With the increase in the value of the electropositivity of metals, the basic nature of their oxides also increases.

NCERT Solutions Class 11 Chemistry Chapter 3 MCQs

Question 14. The increasing order of the ionic radii of the given isoelectronic species is —

1. S^2- , Cl^2- Ca²⁺ , K
2. Ca²⁺ , K⁺ , Cl^– S^2-
3. K⁺ , S^2- , Ca²⁺ , Cl^–
4. Cl-, Ca²⁺, K²⁺, S²⁺

Answer: 2. Ca²⁺ , K⁺ , Cl^– S^2-

For any atom Orion, ionic radius (e/Z) [e=number of electrons and Z= atomic number]. Since the given ions are isoelectronic their electron number is also the same.

It means the higher the atomic number lower the ionic radius of the ion. Therefore, the correct order of the ionic radii ofthe ions is Ca²⁺, K⁺, Cl^– S^2-

Question 15. The first ionization potential of Na is 5.1 eV. The value of electron-gain enthalpy of Na⁺ will be

1. +2.55ev
2. -2.55ev
3. -5.1ev
4. -10.ev

Answer: 3. -5.1ev

For any element, the value of the first ionization potential and that of the electron-gain enthalpy of its unipositive ion. However, for electron gain enthalpy energy is liberated while for ionization potential energy is absorbed. Hence, they have the same magnitude but opposite sign.

Question 16. Which of the given represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se, and Ar

1. Ca<Ba<S<Se<Ar
2. Ca < S < Ba < Se < Ar
3. S < Se < Ca < Ba < Ar
4. Ba < Ca < Se < S < Ar

Answer: 4. Ba < Ca < Se < S < Ar

On moving from left to right across a period in the periodic table, the values of ionization enthalpy of elements increase, while moving down a group it decreases.

Question 17. Ionic radii (A) of N^3- , O^2- and F^–

1. 1.71, 1.40, and 1.36
2. 1.36, 1.40, and 1.71
3. 1.71, 1.36 and 1.40
4. 1.36, 1.71 and 1.40

Answer: 2. 1.36, 1.40 and 1.71

The boiling point of Xe is the highest because the boiling point increases down the group from He to Rn due to an increase in van der Waals force of attraction as the size of the atom increases.

Question 19. Which of the following atoms has the highest first ionization energy

1. Rb
2. Na
3. K
4. Sc

Answer: 4. Sc

Question 20. The group having isoelectronic species is-

1. O^2- , F^– , Na⁺ Mg²⁺
2. O^– , F^– ,Na⁺   Mg²⁺
3. O^2- F^– , Na, Mg²⁺
4. O^– F^–, Na⁺, Mg²⁺

Answer: 1. O^2-, F^–, Na⁺ Mg²⁺

Question 21. Both lithium and magnesium display several similar properties due to the diagonal relationship, however, the incorrect one, is

1. Both form nitrides
2. Nitrates of both li and mg yield nO₂ and O₂ on heating
3. Both form soluble carbonates
4. Both form soluble bicarbonates

Answer: 3. Both form soluble carbonates

Despite having a diagonal relationship, Li and Mg differ in carbonate salts. Magnesium forms basic carbonate [3MgCO₃, Mg(OH)₂, 2H₂O] but lithium like other alkali metals form normal carbonate [Li2CO₃] salts. Hence, O-2, F+, and Mg²⁺ are isoelectronic species.

Question 22. What is the value of electron-gain enthalpy of Na⁺ if IEX

1. -5.1 eV
2. -10.2 eV
3. +2.55 eV
4. +10.2 eV

Answer: 1. -5.1 eV

Class 11 Chemistry Chapter 3 Multiple Choice Questions

Question 23. In which of the following arrangements, the given sequence is not according to the indicated against it

1. HO < H₂S < H₂Se < H₂Te: increasing pKa values
2. NH₃ < PH₃ < ASH₃ < SbH₃: increasing acidity
3. CO₂<SiO₂<SnO₂<PbO₂: increasing oxidising power
4. HF < HC1 < HBr <HI: increasing acidic strength

Answer: 1. HO < H₂S < H₂Se < H₂Te: increasing pKa values

On moving down a group in the periodic table, the covalent character of the hydrides of the elements increases and so their acidity gradually increases. Therefore, the acidic character of the hydrides of the elements of group VA and VILA follows the order— NH₃< PH₃ < AsH₃ < SbH₃ and HF < HCI < HBr < HI On the other hand, the oxidizing power of the oxides in option follows the trend:

CO₂ < SiO₂ < SnO₂ < PbO₂

Question 24. Identify the wrong statement among the following 

1. The atomic radius of the elements increases as one moves down the first group ofthe periodic table
2. The atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table
3. Amongst isoelectronic species, the smaller the positive charge on the cation, the smaller the ionic radius
4. Amongst isoelectronic species, the greater the negative charge on the anion, the larger the the ionic radius

Answer: 3. Amongst isoelectronic species, the smaller the positive charge on the cation, the smaller the ionic radius

On moving from left to right across a period. The proton number and magnitude of the nuclear charge of elements increase gradually leading to a corresponding decrease in the atomic radii of the elements.

Again, moving down a group, with an increase in the atomic number of elements, their atomic radii gradually decrease due to the addition of new shells. For isoelectronic ions, with a decrease in the positive charge of cations and with an increase in the negative charge of anions, ionic radii increase.

Question 25. The reason for lanthanoid contraction is

1. Negligible screening effect of f-orbitals
2. Increasing nuclear charge
3. Decreasingnuclear charge
4. Decreasing screening effect

Answer: 1. Negligible screening effect of f-orbitals.

Question 26. Be²⁺ is isoelectronic with which ofthe following ions?

1. H⁺
2. Li⁺
3. Na⁺
4. Mg2⁺

Answer: 2. Li⁺

⇒ \(\begin{array}{|c|c|c|c|c|c|}
\hline \text { Ion } & \mathrm{Be}^{2+} & \mathrm{H}^{+} & \mathrm{Li}^{+} & \mathrm{Na}^{+} & \mathrm{Mg}^{2+} \\
\hline \text { No. of electron } & 2 & 0 & 2 & 10 & 10 \\
\hline
\end{array}\)

Question 27. Which of the following orders of ionic radii is correctly represented?

1. H→ H+→H
2. F→ O₂→Na⁺
3. Na⁺ → F^–→ O^2-
4. Al³⁺ > Mg²⁺ > N₃

Answer: 2. F→ O₂→Na⁺

None of the options is correct; An atom loses an electron (s) to form a cation. Thus, the radius of the formed cation is less than the parent atom.

Again an atom gains electron (s) to form an anion. So, the radius of the formed anion is greater than the parent atom. Therefore, H^– > H > H⁺. Now, for isoelectronic species, with an increase in their atomic number, their ionic radii decrease. Therefore, the correct orders of ionic radii are O^2-→F¯ > Na+ and N^3- > Mg²⁺ > Al³⁺

Question 28. The species Ar, K⁺, and Ca²⁺ contain the same number of electrons. In which order do their radii increase

1. Ca²⁺ <K+<Ar
2. K+ < Ar < Ca²⁺
3. Ar < K+ < Ca²⁺
4. Ca²⁺ < Ar < K⁺

Answer: 1. Ca²⁺ <K+<Ar

For isoelectronic species atomic/ionic radii decrease with the increase of nuclear charge. Hence increasing order atomic/ionic radii  Ca²⁺ < igK+ < 18Ar.

Periodicity in Properties MCQs Class 11 Chemistry

Question 29. Because of lanthanide contraction, which of the following pairs of elements have nearly the same atomic radii (Numbers in the parenthesis are atomic numbers)

1. Zr(40) and Hf(72)
2. Zr(40) and Ta(73)
3. Ti(22) and Zr(40)
4. Zr(40) and Nb(41)

Answer: 1. Zr(40) and Hf(72)

Zr and Hf have the same atomic radii due to the lanthanide contraction.

Question 30. In which ofthe following options the order of arrangement does not agree with the variation of property indicated against it

1. I < Br < Cl < F (increasing electron-gain enthalpy)
2. Li<Na<K<Rb (increasingmetallic radius)
3. Al³⁺ < Mg^2+_ < Na⁺ < F^–(increasing ionic size)
4. B<C<N<0 (increasing first ionization enthalpy)

Answer: 1. I < Br < Cl < F (increasing electron-gain enthalpy)

The increasing order of negative electron gain enthalpy; I < Br < F < Cl and that of first ionization energy: B < C < O < N.

Question 31. The element Z =114 has been discovered recently. It will belong to which of the following family/groups and electronic configuration—

1. Carbon family, [rn]5/I4⁶d¹⁰7s²7p²
2. Oxygen family, [rn]5/14⁶d¹⁰7s²7p⁴
3. Nitrogen family, [rn]5/14⁶d¹⁰7s²7p⁶
4. Halogen family, [rn]5/I4⁶d¹⁰7s²7p⁵

Answer: 1. Carbon family, [rn]5/i4⁶d¹⁰7s²7p²

The electronic configuration of the element having atomic mass 114 is [Rn]5f¹⁴ 6d¹⁰7s²7p² The outer electronic configuration of the element is the same as that of carbon. So the element should belong to the carbon family.

Question 32. Among CaH₂, BeH₂, BaH₂ the order of ionic character

1. BeH₂<CaH₂<BaH₂
2. CaH₂ < BeH₂ < BaH₂
3. BeH₂ < HaH₂ < CaH₂
4. BaH₂ < BeH₂ < CaH₂

Answer: BeH₂<CaH₂<BaH₂

Down a group metallic character increases. Thus ionic character of the metal hydride increases down the group. Hence least ionic compound is BeH₂.

Question 33. Which ofthe following oxides is most acidic

1. MgO
2. BeO
3. BaO
4. CaO

Answer: 2. BeO

Down a group, the metallic character of the elements increases. Hence down the group basic character of the metallic oxide increases. Thus BeO has the most acidic character. It is an amphoteric oxide whereas others are basic oxides.

Question 34. The first ionization enthalpy of Na, Mg, and Si are 496, 737, and 776 kj.mol^-1 respectively. What will be the first ionization enthalpy potential of a kj mol^-1

1. >766 kj – mol^-1
2. >496 and < 737 k).mol^-1
3. >737 and < 766 kj.mol^-1
4. >496 kj- mol^-1

Answer: 2. >496 and < 737 k).mol^-1

The ionization enthalpy of A1 is lower than that of Mg as the 3p1 electron of A1 is easier to remove than to remove an electron from the fully-filled 3s -orbital of Mg.

Question 35. Which is correct regarding the size of the atom

1. B < Ne
2. Na > K
3. N < O
4. V > Ti

Answer: 2. The atomic radii of noble gases are the largest in their respective periods. This is due to the reason that noble gases have only van der Waals radii.

NCERT Class 11 Chemistry Classification of Elements MCQs

Question 36. An element (X) belongs to the fourth period and fifteenth group of the periodic table. Which one of the following is true regarding the outer electronic configuration of (X)? It has

1. Partially filled d -d-orbitals and filled orbital,
2. Filled s -orbital and filled p orbitals.
3. Filled s -orbital and half-filled p -orbitals.
4. Half-filled d -orbitals and filled s -orbitals.

Answer: 3. filled s -orbital and half-filled p -orbitals.

The electronic configuration of (X) can be written as X: ls²2s²2p⁶3s³3p⁶4s²3d¹⁰4p³ So, element (X) has filled s -and d -orbitals and half-filled p -orbitals.

Question 37. Which of the following transformations produces the maximum amount of energy

1. M-(g)→M(g)
2. M(g)→M-(g)
3. M⁺(g)→M³⁺(g)

Answer: 4. M²⁺+(g)→M³⁺(g)

Question 38. The amount of energy released when 106 atoms of iodine in a vapor state are converted into- ions is 4.8 x 10-13 J. What is the electron affinity of iodine in kJ mol^-1

1. 489
2. 289
3. 259
4. 389

Answer: 2. 289

Question 39. The elements that occupy the peaks of the ionization energy potential curve are

1. Na, K, Rb, Cs
2. Cl, F, Br, I
3. Na, Mg, Cl, I
4. He, Ne, Ar, Kr

Answer: 4. He, Ne, Ar, Kr

Question 40. The electronic configuration ofthe atom having maximum difference second and third ionization enthalpies is

1. ls²2s²2p⁶3s²
2. ls²2s²2p⁶3s²3p¹
3. ls²2s²2p⁶3s²3p²
4. ls²2s²2p⁶3s¹

Answer: 1. ls²2s²2p⁶3s²

Question 41. Identify the least stable ion amongst the following

1. Li^–
2. Be^–
3. B^–
4. C^–

Answer: 2. Be^–

Question 42. If each orbital can accommodate a maximum of four electrons, the number of elements in the third period of the periodic table will be

1. 10
2. 12
3. 14
4. 16

Answer: 4. 16

Question 43. Three elements X, Y, and Z are present in the third short period and their oxides are ionic, amphoteric, and giant molecules respectively. The correct order of atomic numbers of X, Y, and Z is

1. Z< Y<X
2. X<Z<Y
3. Y<Z<X
4. X<Y<Z

Answer: 4. X<Y<Z

Question 44. A gaseous mixture of He, Ne, Ar, and Kr is irradiated with photons of frequency appropriate to ionize Ar. The ion(s) present in the mixture will be

1. Only
2. Ar+ and He+
3. Ar+ and Ne
4. Ar+ and Kr+

Answer: 4. Ar+ and Kr+

MCQs on Classification of Elements Chapter 3 Class 11 Chemistry

Question 45. Boiling point of Kr & Rn are -152°C & -62°C respectively. Then the boiling point of Xe is expected to be

1. -92C
2. -87C
3. -107C
4. 77C

Answer: 3. -107C

Question 46. Which ofthe following is smallest in size

1. Li+ (aq)
2. Na+ (aq)
3. K+ (aq)
4. Rb+ (aq)

Answer: 4. Rb+ (aq)

Question 47. The atomic radius of Li is 1.23 A and the ionic radius of Li+ is 0.76 A. Percentage of the volume occupied by a single valence electron in Li is

1. 35
2. 52.5
3. 76.4
4. 83.72

Answer: 3. 76.4

Question 48. The number of valence electrons in element A is 3 and that in element B is 6. The most probable compound from A and B is

1. A₂B
2. AB₂
3. A₆B₃
4. A₂B₃

Answer: 4. A₂B₃

Class 11 Chemistry Periodicity in Properties Multiple Choice Questions

Question 49. The ionic radius of ‘Cr’ is the minimum in which of the following compounds

1. K₂CrO₄
2. CrF₃
3. CrO₂
4. CrCl₃

Answer: 1. K₂CrO₄

Question 50. The correct order of radii is

1. N < Be < B
2. F^– <O^2- <N^3-
3. Na < Li < K
4. Fe³⁺ <Fe²⁺<Fe⁴⁺

Answer: 2. F^– <O^2- <N^3-